Trigonometry — Revision Notes

The question	Where it is answered	The tool
What is sin 130°? Why is cos 130° negative?	9.1 Obtuse angles	the extended definitions, and sin `A` = sin (180° − `A`)
What is the area of a triangle that has no obvious height?	9.1 Area of a triangle	½`ab` sin `C`, with `C` the angle between the two sides
The question gives two angles and a side, or two sides and an angle that is not between them.	9.2 The Sine Rule	the Sine Rule — pair each side with the angle opposite it
The question gives three sides, or two sides and the angle between them.	9.3 The Cosine Rule	the Cosine Rule — the only rule that copes with no side–angle pair
How tall is it, how far is it, and on what bearing? What about inside a box or a pyramid?	9.4 Elevation & bearings, 9.5 Three dimensions	a clear diagram, then whichever of the above the triangle in it needs

What you are standing on ⓘ Recap

Two results carry over from Secondary Two, and this chapter uses both constantly.

In a right-angled triangle, with θ one of the two acute angles, sin θ = opposite ÷ hypotenuse, cos θ = adjacent ÷ hypotenuse and tan θ = opposite ÷ adjacent. Pythagoras' Theorem says the square on the hypotenuse equals the sum of the squares on the other two sides.

Both of those need the right angle. They apply to a triangle that has one, and to no other. That is why the rest of this chapter exists: most triangles do not have one.

Where this comes from

There are two ways forward once a triangle has no right angle, and a good solution often uses both: cut the triangle into right-angled pieces with a perpendicular, or use a rule — sine or cosine — that has already done that cutting for you. The derivations in 9.2 and 9.3 are the second route being built out of the first.

Past-paper questions

G3 N2021 · Paper 1 · Q2 2 marks

In the triangle, AC = 20.5 cm, BC = 13.3 cm and angle ABC = 90°. Calculate AB. [2]

Worked solution

The right angle is at B, so the side facing it, AC, is the hypotenuse. By Pythagoras' Theorem

{A B}^{2} + {B C}^{2} = {A C}^{2}

.

${A B}^{2} = {20.5}^{2} - {13.3}^{2} = 420.25 - 176.89 = 243.36$

$A B = \sqrt{243.36}$ = 15.6 cm (exactly).

Why this works. The hypotenuse faces the right angle, so here it is AC and the line the marks sit on is ${A B}^{2} = {20.5}^{2} - {13.3}^{2}$ , not ${20.5}^{2} + {13.3}^{2}$ . Pythagoras adds the two short sides, so with the hypotenuse known the working subtracts. The answer is exact — 243.36 is ${15.6}^{2}$ — so no rounding statement is needed.

G3 N2022 · Paper 1 · Q22 3 marks

The diagram shows a trapezium ABCD. Angle ABD = angle BDC = 90°. AD = 12.8 cm, BC = 10.3 cm and angle DAB = 38°. Calculate angle CBD. [3]

Worked solution

In right-angled triangle ABD, BD is opposite the 38° and AD = 12.8 is the hypotenuse:

$B D = 12.8 \sin 38° = 7.8804…$ cm

In right-angled triangle BDC the right angle is at D, so BC = 10.3 is the hypotenuse and BD is adjacent to the angle CBD:

$\cos ∠ C B D = \frac{B D}{10.3} = \frac{7.8804…}{10.3} = 0.76509…$

Angle CBD = 40.084…° = 40.1° (to 1 d.p.).

Why this works. No rule to choose here: both triangles are right-angled, so this is SOH–CAH–TOA used twice with the shared side BD as the bridge. The marks sit on that bridge, so store 7.8804… and use it — a two-figure 7.9 turns 40.1° into 39.9°. In the second triangle the right angle is at D, so BC is the hypotenuse and BD is adjacent: cosine.

G3 N2022 · Paper 2 · Q6 12 marks

The diagram shows two circles that touch at C. A, B and C are points on the smaller circle, centre O. C, D, E and F are points on the larger circle, centre P. AOCPE, BCD and DPF are straight lines. Angle CPF = 96°.

Find angle DEP. [2]
Show that triangle ABC is similar to triangle FCD. Give a reason for each statement you make. [3]
DE = 7.21 cm, DF = 9.70 cm and BD = 9.10 cm and angle CPF = 96°.
1. Calculate AB. [4]
2. Calculate the length of the minor arc AB. [3]

Worked solution

(a) DPF is a straight line, so angle DPC = 180° − 96° = 84°. That is the angle at the centre standing on arc CD, and angle DEC stands on the same arc at the circumference, so

angle DEP = angle DEC = 84° ÷ 2 = 42° (P lies on CE).

(b) AC is a diameter of the smaller circle and DF is a diameter of the larger, so

angle ABC = angle FCD = 90° (angle in a semicircle).

PD = PC (radii), so triangle PDC is isosceles with apex angle DPC = 84°, giving angle PDC = angle PCD = 48° (base angles of an isosceles triangle). Hence angle DCE = 48°, and

angle ACB = angle DCE = 48° (vertically opposite angles, since BCD and ACE are straight lines) = angle FDC.

Two pairs of equal angles, so triangle ABC is similar to triangle FCD (AA).

(c)(i) In triangle FCD, right-angled at C, with hypotenuse DF = 9.70 and angle FDC = 48°:

$C D = 9.70 \cos 48° = 6.4905…$ cm, so $B C = B D - C D = 9.10 - 6.4905… = 2.6094…$ cm.

Triangle ABC is right-angled at B with angle ACB = 48°, so

$A B = B C \times \tan 48° = 2.8980…$ = 2.90 cm (to 3 s.f.).

(c)(ii) In the same triangle, $A C = \frac{B C}{\cos 48°} = 3.8997…$ cm, and AC is a diameter, so the smaller circle has radius 1.94987… cm.

Angle AOB = 2 × angle ACB = 96° (angle at the centre is twice the angle at the circumference), so

arc AB = $\frac{96}{360} \times 2 π \times 1.94987… = 3.2670…$ = 3.27 cm (to 3 s.f.).

Why this works. Only the last four marks are trigonometry; the rest is angle-chasing, and the link that carries it is the isosceles triangle PDC, whose two radii turn the 84° at P into two 48° base angles. That 48° is the number every later part runs on, and the similarity transfers it into the small triangle. In (c)(ii), "minor arc" means 96°, not 360 − 96.

The lettering every formula in this chapter assumes

The sine rule, the cosine rule and the area formula are all printed using one convention. Setting that convention up on your own sketch is what makes the printed version usable.

In triangle ABC, the capital letters A, B and C name the three vertices, and also the three angles at them. The matching small letters a, b and c name the lengths of the sides opposite those vertices — so a is the side BC, b is the side CA, and c is the side AB. A side is never next to the letter it is named after.

Fig. 9.1

Each colour links an angle to the side opposite it. The side a is the one you can see across the triangle from A; it is the only side that does not touch A. Reading a question, your first job is to decide which letters you have been given — and the very first thing to write down is a labelled sketch.

Real questions rarely use the letters A, B and C. The convention is about the pattern, not the letters: in triangle PQR, the side opposite P is QR.

More detail

Questions use PQR, or XYZ, or place names. Follow the pattern and the printed rule transfers straight across: in triangle PQR the sine rule reads QR / sin P = PR / sin Q = PQ / sin R. Writing the rule out in the question's own letters, before substituting anything, is the habit this chapter rewards most.

Sine and cosine of an obtuse angle

An angle in a triangle runs from just above 0° to just below 180°, so sine and cosine have to be extended past the acute angles the right-angled definitions cover. The extension is done with coordinates.

For any angle A with 0° ≤ A ≤ 180°, taking P(x, y) on a circle of radius r centred at the origin,

\sin A = \frac{y}{r}, \cos A = \frac{x}{r}, \tan A = \frac{y}{x}

For an acute A this agrees exactly with opposite ÷ hypotenuse and adjacent ÷ hypotenuse, so nothing you already knew has changed.

Fig. 9.2

Draw a circle of radius r centred at the origin O, and let P(x, y) be the point where the arm of the angle A cuts it. For an obtuse A the point lands in the second quadrant, so y is still positive but x is negative.

The signs fall straight out of the picture. The radius r is a length, so it is positive; and P stays on or above the x-axis in this range, so y is never negative.

	`A` is acute (0° to 90°)	`A` is obtuse (90° to 180°)
sin `A`	positive	positive — `y` is still above the axis
cos `A`	positive	negative — `x` has crossed to the left

A negative cosine is the calculator saying the angle is obtuse. A positive sine says nothing about which it is, and that one asymmetry is behind every difficulty in the rest of the chapter.

\sin A = \sin (180° - A)

\cos A = - \cos (180° - A)

Fig. 9.3

Reflecting P(x, y) in the y-axis gives P′(−x, y), which is the point for the angle 180° − A. The y-coordinate is unchanged and the x-coordinate has only changed sign.

One sentence holds both: supplementary angles have the same sine and opposite cosines. Neither line is printed on the paper.

Walkthrough 1 — values without a calculator Basic

Given that sin 34° = 0.5592 and cos 127° = −0.6018, write down, without using a calculator, the value of (i) sin 146° and (ii) cos 53°.

180° − 146° = 34°

146° is obtuse, so it is not a value you have been given directly. The one relationship you have connects an angle to 180° − itself, so that subtraction is always the opening move — and here it lands on 34°, which is given.
sin 146° = sin 34° = 0.5592

Supplementary angles have the same sine, so no sign change happens. The answer is positive, exactly as the sign table demands for any angle between 0° and 180°.
180° − 53° = 127°

The same move in the other direction: 53° is acute and is not given, but its supplement 127° is. Subtracting from 180° is a two-way street.
cos 53° = −cos 127° = −(−0.6018) = 0.6018

Supplementary angles have opposite cosines, so the minus sign in the rule cancels the minus sign in the given value. The result is positive, and it has to be: 53° is acute, and the cosine of an acute angle is never negative.

Walkthrough 2 — going backwards, from a value to an angle Basic

Find all the values of x between 0 and 180 for which (i) sin x° = 0.72 and (ii) cos x° = −0.41, giving your answers correct to 1 decimal place.

x = sin⁻¹ 0.72 = 46.054…

The calculator's inverse sine only ever returns the acute answer, because that is the one it was built to give. Keep the extra decimals for now; the rounding is a separate step at the end.
or x = 180 − 46.054… = 133.945…

Because sin (180° − A) = sin A, the supplement has exactly the same sine, so it is a second solution — and it is still inside the range 0 to 180, so it counts.
x = 46.1 or x = 133.9

Both roots, rounded once, at the end. The question asks for all the values, so the obtuse answer is written down beside the acute one.
x = cos⁻¹ (−0.41) = 114.204…

A negative cosine, so this angle is obtuse — and here the calculator hands you the obtuse value straight away, because inverse cosine returns everything from 0° to 180°.
x = 114.2 only

There is no second answer this time. Different angles between 0° and 180° always have different cosines — the sign changes partway through the range, which is exactly what stops two of them from matching. Cosine pins an angle down; sine does not.

Where this comes from

Put the acute angle A and the obtuse angle 180° − A on the same circle and the relationship becomes obvious: the two points are mirror images in the y-axis, so they sit at the same height but at opposite distances from it, which is exactly what the reflection figure shows.

The tangent follows from the pair — tan A = −tan (180° − A) — but outcome 4.4 says "sine and cosine", and an exam question will not ask for the tangent of an obtuse angle. It is worth knowing only so that a negative tangent on your calculator is not a surprise.

Past-paper questions

G3 N2022 · Paper 1 · Q19 2 marks

Given that 5 sin x = 2, find the two possible values for angle x, where 0° ≤ x ≤ 180°. [2]

Worked solution

$\sin x = \frac{2}{5} = 0.4$

The calculator gives the acute angle: x = sin⁻¹ 0.4 = 23.578…°

The obtuse angle with the same sine is 180° − 23.578…° = 156.421…°.

x = 23.6° or x = 156.4° (to 1 d.p.).

Why this works. The word two and the range up to 180° are the whole question: a calculator returns only the acute answer. The second mark is for sin (180° − θ) = sin θ, so the obtuse partner comes from subtracting from 180° — not from adding 180°, and not from 360° − θ. Check both lie inside the stated range.

G3 N2023 · Paper 1 · Q2(a) 2 marks

sin x° = 0.9301
Find two possible values of x in the range 0 ≤ x ≤ 180. [2]

Worked solution

sin⁻¹ 0.9301 = 68.450…

The other angle in the range has the same sine: 180 − 68.450… = 111.549…

x = 68.5 or x = 111.5 (to 1 d.p.).

Why this works. One extra detail is worth noticing: the degree sign is printed on the x°, not on the range, so x is a number and the answer is written without a degree sign. Sine is positive right across 0° to 180°, so a value like 0.9301 has exactly two solutions, one either side of 90°. Check: 68.5 and 111.5 sit the same distance from 90.

G3 N2024 · Paper 1 · Q20 2 marks

sin (2x°) = 0.561
Find two possible values of x in the range 0 ≤ x ≤ 90. [2]

Worked solution

Solve for 2x first. Since 0 ≤ x ≤ 90, the angle 2x runs over 0 ≤ 2x ≤ 180, so there are two solutions:

2x = sin⁻¹ 0.561 = 34.124… or 2x = 180 − 34.124… = 145.875…

Halving each: x = 17.1 or x = 72.9 (to 1 d.p.).

Why this works. The equation is about 2x while the range is given for x, so translate the range first: 0 to 90 for x is 0 to 180 for 2x. Collect both values of 2x and halve only at the end — halving first sends the second answer to 162.9, outside the range. Keep 34.124… unrounded before subtracting.

Area of a triangle — ½`ab` sin `C`

½ × base × height needs a height, and a triangle given by two sides and the angle between them does not come with one. This formula gets the area without ever finding it.

Area of △ A B C = \frac{1}{2} a b \sin C

and, since the same argument works from any vertex,

Area = \frac{1}{2} b c \sin A = \frac{1}{2} a c \sin B

The angle is the one between the two sides. The letters say so: ½ab sin C uses the sides a and b, which are the two that meet at C. Any other angle gives a number that is not the area.

An obtuse included angle is not a special case. Because sin C stays positive right up to 180°, the formula never produces a negative area and never needs adjusting. Feed 118° straight in.

Walkthrough 3 — area from two sides and the angle between them Basic

In triangle PQR, PQ = 9.4 cm, QR = 7.2 cm and ∠PQR = 118°. Find the area of triangle PQR.

Area = ½ × PQ × QR × sin ∠PQR

118° is the angle at Q, and PQ and QR are the two sides meeting there, so the given angle is the included one. Writing the formula in the question's own letters is what makes that visible.
= ½ × 9.4 × 7.2 × sin 118°

Substitute. The 118° goes in exactly as it stands: sin 118° is a perfectly ordinary positive number, and nothing needs converting to an acute angle first.
= 33.84 × 0.88294… = 29.878…

½ × 9.4 × 7.2 = 33.84, and the calculator supplies sin 118°. Keep the digits: this is an intermediate value, not the answer.
= 29.9 cm² (to 3 s.f.)

Round once, at the end, to the paper's default accuracy, and write cm² — an area carries squared units even when every length in the question was in plain centimetres.

Walkthrough 4 — the area is given, a length is not Intermediate

In triangle ABC, AB = 4x cm, AC = 7x cm and ∠BAC = 106°. Given that the area of the triangle is 84 cm², find the value of x.

½ × 4x × 7x × sin 106° = 84

Same formula, read the other way round: the area is the known quantity and one of the lengths is not. 106° is again the included angle, sitting between the two sides named.
14x² sin 106° = 84

½ × 4 × 7 = 14, and x × x = x². Collecting the numbers before touching the calculator keeps the algebra visible and the arithmetic short.
$x^{2} = \frac{84}{14 \sin 106°} = 6.2417…$

Divide by the whole coefficient 14 sin 106° in one step, on the calculator, rather than working out sin 106° and then dividing twice. Fewer keystrokes, fewer rounding errors.
x = $\sqrt{6.2417…}$ = 2.4983…

Only the positive square root is taken. The negative root is a genuine solution of the equation but not of the problem: 4x is a length, and lengths cannot be negative.
x = 2.50 (to 3 s.f.)

The answer is a plain number, not a length, so it carries no units — but it is still non-exact, so it still goes to 3 significant figures. Write the final zero: 2.50 shows three figures, 2.5 shows two.

Walkthrough 5 — the area is given, the angle is not Intermediate

In triangle XYZ, XY = 11.5 cm and YZ = 8.4 cm. The area of triangle XYZ is 37.2 cm². Find the two possible values of ∠XYZ, giving your answers correct to 1 decimal place.

½ × 11.5 × 8.4 × sin ∠XYZ = 37.2

∠XYZ is the angle at Y, and XY and YZ are the two sides meeting at Y, so the area formula applies with the unknown in the sine.
48.3 sin ∠XYZ = 37.2

½ × 11.5 × 8.4 = 48.3. Simplifying the numerical coefficient first turns the equation into the plainest possible shape: a number times a sine equals a number.
sin ∠XYZ = $\frac{37.2}{48.3}$ = 0.77018…

A sine, not an angle. The value is between 0 and 1, so a triangle like this really does exist — a value bigger than 1 would have been the signal that no such triangle can be drawn.
∠XYZ = sin⁻¹ 0.77018… = 50.370…° = 50.4°

The calculator returns the acute angle. Round to 1 decimal place, which is what the paper asks for angles in degrees unless it says otherwise.
or ∠XYZ = 180° − 50.370…° = 129.6°

The supplement has the same sine, so it gives the same area: opening the angle at Y out to 129.6° builds a different triangle of equal area. Take the supplement of the unrounded value, or the second answer lands 0.1° out.

Where this comes from

Drop a perpendicular from a vertex and measure it with the sine ratio: that is the whole derivation, and it works whether the included angle is acute or obtuse.

Fig. 9.4

Taking BC as the base, the height is AD. In the right-angled triangle ACD, h = b sin C, so the area is ½ × a × b sin C.

Fig. 9.5

If C is obtuse the foot D falls outside the base, and h = b sin (180° − C). But that is the same as b sin C, so the formula is unchanged.

The formula is printed on the paper, word for word as "Area of triangle ABC = ½ab sin C". What the printed line cannot do is tell you which of your sides is a and which of your angles is C; that reading is where the marks are.

Check yourself

Given that cos 68° = 0.3746, write down the value of cos 112°.
Answer
180° − 112° = 68°, and supplementary angles have opposite cosines, so cos 112° = −cos 68° = −0.3746.
Solve sin x° = 0.3, giving all the values of x between 0 and 180, correct to 1 decimal place.
Answer
sin⁻¹ 0.3 = 17.457…, and 180 − 17.457… = 162.542…, so x = 17.5 or x = 162.5. Both are in range.
Explain why cos x° = 0.85 has only one solution between 0 and 180, while sin x° = 0.85 has two.
Answer
Across 0° to 180° the cosine decreases all the way, from 1 down to −1, so it never takes the same value twice. The sine climbs from 0 to 1 and then falls back to 0, so every value strictly between 0 and 1 is reached twice — once at an acute angle and once at its supplement.
Find the area of triangle LMN, where LM = 6.3 cm, LN = 10.2 cm and ∠MLN = 143°.
Answer
The angle is at L, between the two given sides, so area = ½ × 6.3 × 10.2 × sin 143° = 32.13 × 0.60181… = 19.336… = 19.3 cm² (to 3 s.f.).
In triangle DEF, DE = 9 cm, EF = 12 cm and the area is 40 cm². Find the two possible values of ∠DEF.
Answer
½ × 9 × 12 × sin ∠DEF = 40, so sin ∠DEF = 40 ÷ 54 = 0.74074… and ∠DEF = 47.794…° or 180° − 47.794…°, giving 47.8° or 132.2°.
A student writes "area = ½ × 8 × 5 × sin 40°" for a triangle in which the sides of 8 cm and 5 cm meet at 65° and the 40° angle is at another vertex. What has gone wrong?
Answer
The formula needs the included angle — the one between the two sides used. Here the two sides meet at 65°, so the working should read ½ × 8 × 5 × sin 65°. Using an angle from elsewhere in the triangle gives a number, but not the area.

Past-paper questions

G3 N2016 · Paper 1 · Q6 2 marks

The area of triangle ABC is 58.6 cm². AB = 18.7 cm and BC = 12.8 cm. Find the two possible sizes of angle ABC. [2]

Worked solution

The two given sides meet at B, so the angle between them is ABC and the area formula applies directly:

$\frac{1}{2} \times 18.7 \times 12.8 \times \sin ∠ A B C = 58.6$

$\sin ∠ A B C = \frac{58.6}{119.68} = 0.489639…$

Angle ABC = 29.316…° or 180° − 29.316…° = 150.683…°, so

angle ABC = 29.3° or 150.7° (to 1 d.p.).

Why this works. The formula is printed, so the first mark is for the lettering: the angle is the one between the two given sides, and AB and BC both touch B. The second is for the word two — a sine below 1 comes from an acute angle and its obtuse partner. Both survive because nothing else fixes a third side or angle.

G3 N2018 · Paper 2 · Q7(b)(i) 3 marks

AB is a diameter of the large circle, centre O. CD is a diameter of the small circle, centre O. AC and BD are tangents to the small circle. The radius of the large circle is 7 cm and angle OAC = 30°. Calculate the area of triangle OAC. [3]

Worked solution

AC touches the small circle at C, and a tangent is perpendicular to the radius at the point of contact, so angle OCA = 90°. In right-angled triangle OAC the hypotenuse is OA = 7 cm.

$O C = 7 \sin 30° = 3.5$ cm and $A C = 7 \cos 30° = \sqrt{36.75} = 6.0621…$ cm

Area of triangle OAC $= \frac{1}{2} \times 3.5 \times 6.0621… = 10.6088…$ = 10.6 cm² (to 3 s.f.).

Why this works. The circle supplies one fact, through the word tangent: the tangent is perpendicular to the radius at contact, so angle OCA = 90°, with that reason written. A one-line route earns the same marks: area = $\frac{1}{2} \times 7 \times 6.0621… \times \sin 30°$ . OA is a hypotenuse, not a leg, so it pairs with the sine formula.

The Sine Rule

The first of the two rules that work in any triangle. It pairs each side with the angle facing it.

In any triangle ABC,

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

or, turned upside down,

\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}

Use the version that puts the unknown on top. Finding a side, use the first form; finding an angle, use the second. Choosing before you substitute saves a line of algebra.

Two of the three fractions at a time. Pick the pair holding the unknown and one complete side–angle pair, and write only those two — a line with three fractions in it cannot be solved.

Where this comes from

Drop a perpendicular from B to AC in any triangle and call its length h. Looking at the right-angled triangle on the left, h = c sin A; looking at the one on the right, h = a sin C. Those are the same h, so c sin A = a sin C, which rearranges into a statement about ratios — and repeating the argument from another vertex extends it to all three.

Both forms say the same thing: the first needs one rearrangement to find a side and two to find an angle, and the second is the other way round.

The rule is printed on the paper, as a / sin A = b / sin B = c / sin C — in the a, b, c lettering and nothing else. It will not tell you that in triangle PQR the side opposite P is QR; set the lettering up yourself, on your own sketch, every time.

When the sine rule is the right tool

Every fraction in the sine rule is a side over the sine of the angle opposite it. So the rule is usable exactly when the question hands you one complete pair — a side and the angle facing it — plus one more piece of information.

Use the Sine Rule when you are given (1) two angles and the length of one side, or (2) the lengths of two sides and one non-included angle — an angle that is not between them.

Case (1) is safe; case (2) needs a decision. Two sides and a non-included angle leave you holding a sine, and a sine does not separate an angle from its supplement. That decision is the ambiguous case.

Walkthrough 6 — two angles and a side Basic

In triangle ABC, ∠BAC = 71.4°, ∠ABC = 52.8° and AB = 9.3 cm. Find (i) ∠ACB and (ii) the length of BC.

∠ACB = 180° − 71.4° − 52.8° = 55.8°

Angle sum of a triangle. This has to come first: the sine rule needs a side paired with the angle opposite it, and the only side you have is AB, whose opposite angle is the one at C.
$\frac{B C}{\sin A} = \frac{A B}{\sin C}$

Two fractions, not three, and the unknown is on top because a length is wanted. BC is opposite A and AB is opposite C, so this pairing is the one the rule allows.
BC = $\frac{9.3 \sin 71.4°}{\sin 55.8°}$ = 10.657…

Multiplying both sides by sin 71.4° is the only rearrangement needed, which is exactly why the unknown was put on top. Type the whole fraction in one go and keep the extra digits.
BC = 10.7 cm (to 3 s.f.)

A length, so 3 significant figures and centimetres. A quick sanity check: BC faces the biggest angle (71.4°) so it should be the longest side, and 10.7 cm is indeed longer than the 9.3 cm given.

Walkthrough 7 — two sides and a non-included angle Basic

In triangle PQR, ∠PQR = 43°, PQ = 9.4 cm and PR = 12.6 cm. Find ∠PRQ.

Fig. 9.6

$\frac{\sin R}{P Q} = \frac{\sin Q}{P R}$

The 43° at Q faces PR, so that is the complete pair; the angle wanted is at R, which faces PQ. The sines are on top this time because an angle is the unknown.
sin R = $\frac{9.4 \sin 43°}{12.6}$ = 0.50879…

One multiplication clears the fraction. The result is a sine, not an angle — and because it is comfortably less than 1, a triangle with these measurements does exist.
R = sin⁻¹ 0.50879… = 30.583…° or 180° − 30.583…° = 149.416…°

Both are written down, because both really do have that sine. The calculator only offers the acute one, so the obtuse alternative has to be produced by hand — and then tested.
If R were obtuse: 43° + 149.416…° = 192.4° > 180°, which is impossible.

Two angles of a triangle can never add to more than 180°, so the obtuse option is rejected — and the rejection is written out, because showing why the other root fails is exactly what the mark is for.
∠PRQ = 30.6° (to 1 d.p.)

One answer survives, so the question is not ambiguous after all. Angles in degrees go to 1 decimal place, and the rounding happens only now.

More detail

With two angles you can always find the third from the angle sum, so every side is then reachable and there is no doubt anywhere in case (1).

In case (2) the decision has to be made in writing: say whether the obtuse alternative builds a real triangle or not. That written decision carries marks of its own, and the next section is about how to make it.

Past-paper questions

G3 N2021 · Paper 1 · Q19 4 marks

In the triangle, AC = 15.6 cm, angle CAB = 82° and angle ABC = 56°. Calculate the area of the triangle. [4]

Worked solution

Only one side is known, so a second one has to be built before any area formula can be used. Angle sum first:

angle ACB = 180° − 82° − 56° = 42°

Now 15.6 faces the 56°, and AB faces the 42°, so the sine rule pairs them:

$\frac{A B}{\sin 42°} = \frac{15.6}{\sin 56°}$ , so $A B = \frac{15.6 \sin 42°}{\sin 56°} = 12.5910…$ cm

The 82° sits between AC and AB, so

area $= \frac{1}{2} \times 15.6 \times 12.5910… \times \sin 82° = 97.254…$ = 97.3 cm² (to 3 s.f.).

Why this works. Four marks for an area signals that ½ab sin C comes last: it needs two sides with the angle between them, and only one side is given. The angle sum supplies the third angle, which completes a side–angle pair for the sine rule and builds the second side. Carry the unrounded 12.5910… into the last line: 12.59 gives 97.246, which prints as 97.2.

The ambiguous case — when both answers are real

In Walkthrough 7 the obtuse option was killed by the angle sum. It does not always die, and deciding which happens is worth marks of its own.

The ambiguous case is a set of measurements that fits two different triangles. It can only arise from "two sides and a non-included angle", because that is the only case in which the sine rule leaves you holding a sine rather than an angle.

Fig. 9.7

Both triangles ABC₁ and ABC₂ have the same angle B, the same side c and the same side b — and yet they are different triangles. The angle at C₁ is acute; the angle at C₂ is its supplement, and obtuse.

The test is one line of arithmetic, and it is written down. Add the supplement of the acute angle to the given angle.
• Over 180° — reject the obtuse option, give one answer.
• Under 180° — give both answers.

The question usually says which case it is in. "Find the two possible values", or a diagram with an obviously obtuse angle, keeps both; "the triangle is acute-angled", or a diagram that is clearly not, leaves one.

Walkthrough 8 — both triangles are real Intermediate

In triangle ABC, ∠ABC = 40°, AB = 15.2 cm and AC = 11.8 cm. Find the two possible values of ∠ACB and the corresponding lengths of BC.

Fig. 9.8

sin ∠ACB = $\frac{15.2 \sin 40°}{11.8}$ = 0.82799…

The sine rule with the sines on top, using the pair 40° opposite AC and the unknown angle at C opposite AB. Two sides and an angle not between them — the only arrangement that can be ambiguous.
∠ACB = 55.893…° or 180° − 55.893…° = 124.106…°

Both candidates written out in full. Neither is discarded yet; whether an obtuse angle at C is possible is a question about the angle sum, not about the calculator.
40° + 124.106…° = 164.1° < 180°, so both are possible: ∠ACB = 55.9° or 124.1° (to 1 d.p.)

The obtuse option leaves 15.9° over for the angle at A, which is a perfectly good triangle. Because the test passes, nothing is rejected and the answer genuinely has two branches — exactly what the question's word "two" promised.
Case 1: ∠BAC = 180° − 40° − 55.893…° = 84.106…°, so $B C = \frac{11.8 \sin 84.106°}{\sin 40°} = 18.260…$

Each branch is finished separately, and each starts by finding the third angle. BC is opposite A, and AC is opposite B, so those two make the pair. The unrounded 55.893… is used, not 55.9.
Case 2: ∠BAC = 180° − 40° − 124.106…° = 15.893…°, so $B C = \frac{11.8 \sin 15.893°}{\sin 40°} = 5.0272…$

The same three lines with the other angle. A much smaller angle at A gives a much shorter BC, which is what the picture shows: C₂ sits far nearer to B.
∠ACB = 55.9° with BC = 18.3 cm, or ∠ACB = 124.1° with BC = 5.03 cm

Pair each angle with its own length — the two results describe two different triangles and must not be mixed. Angles to 1 d.p., lengths to 3 s.f., and the rounding done once, right at the end.

Where this comes from

Fix an angle at B and a side c; then swinging a side of length b down from A can meet the other arm of the angle at two different points, which is what the figure above shows.

When both survive the test, each branch is a whole triangle: give both answers together with everything each of them implies — the third angle, and any length the question asks for, worked out separately for each.

Check yourself

In triangle ABC, ∠BAC = 38°, ∠ABC = 96° and AB = 7.5 cm. Find AC.
Answer
∠ACB = 180° − 38° − 96° = 46°. Then AC ÷ sin 96° = 7.5 ÷ sin 46°, so AC = 7.5 sin 96° ÷ sin 46° = 10.369… = 10.4 cm (to 3 s.f.).
In triangle PQR, ∠QPR = 62°, QR = 14 cm and PR = 10 cm. Find ∠PQR, explaining why there is only one answer.
Answer
sin ∠PQR = 10 sin 62° ÷ 14 = 0.63067…, so ∠PQR = 39.100…° or 140.899…°. But 62° + 140.899…° = 202.9° > 180°, so the obtuse value is impossible and ∠PQR = 39.1°.
In triangle XYZ, ∠XYZ = 32°, XY = 18 cm and XZ = 11 cm. Find the two possible values of ∠XZY.
Answer
sin ∠XZY = 18 sin 32° ÷ 11 = 0.86714…, so ∠XZY = 60.128…° or 119.871…°. Since 32° + 119.871…° = 151.9° < 180°, both work: 60.1° or 119.9°.
Why can the sine rule never be used on a triangle in which you are told all three sides?
Answer
Every fraction in the sine rule contains an angle. With three sides and no angle at all there is no complete side-and-opposite-angle pair to set the unknown fraction equal to, so the rule gives an equation with two unknowns in it. That case belongs to the cosine rule.
A student finds sin θ = 1.34 from the sine rule. What does that tell them?
Answer
No angle has a sine bigger than 1, so no such triangle exists — or, far more likely in an exam, a side has been paired with the wrong angle. Go back and check that every fraction holds a side over the sine of the angle opposite it.

The Cosine Rule

Three sides, or two sides and the angle squeezed between them: in both of those there is no side paired with the angle facing it, so the sine rule cannot start. The cosine rule fills exactly that gap.

In any triangle ABC,

a^{2} = b^{2} + c^{2} - 2 b c \cos A

and, rearranged to give the angle,

\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}

The same statement can be started from any vertex: b² = a² + c² − 2ac cos B and c² = a² + b² − 2ab cos C.

Read the pattern, not the letters. The square of the side opposite the angle equals the sum of the squares of the other two, minus twice their product times the cosine of that angle. Side a pairs with angle A.

Only the first form is printed on the paper. The rearranged version is derived or memorised, and the square that is subtracted is the one opposite the angle you want.

Where this comes from

The cosine rule comes out of the same perpendicular as the sine rule, used differently.

Fig. 9.9

With BD = h and AD = x, Pythagoras in triangle BDC gives a² = h² + (b − x)², and in triangle ABD gives c² = h² + x². Since x = c cos A, eliminating h and x leaves a² = b² + c² − 2bc cos A.

Deriving the rearranged form takes two lines and is safer than half-remembering which square is subtracted — it is always the one opposite the angle you want.

Pythagoras' Theorem is a special case. Put A = 90° and cos A = 0, so the last term disappears and a² = b² + c². The −2bc cos A term is the correction for the angle not being a right angle: negative when A is acute (shrinking a) and positive when A is obtuse (stretching it), because cos A changes sign at 90°.

Finding a side: two sides and the angle between them

Walkthrough 9 — the third side Basic

In triangle ABC, AB = 13.6 cm, BC = 9.5 cm and ∠ABC = 104°. Calculate the length of AC.

AC² = AB² + BC² − 2 × AB × BC × cos ∠ABC

104° sits between the two given sides, so there is no complete side–angle pair and the sine rule cannot start. The cosine rule is written with AC alone on the left because AC is the side opposite the given angle.
= 13.6² + 9.5² − 2 × 13.6 × 9.5 × cos 104°

Straight substitution, with the angle left as 104°. Nothing needs converting: the calculator handles the cosine of an obtuse angle by itself.
= 184.96 + 90.25 − 258.4 × (−0.24192…) = 337.72…

cos 104° is negative, so subtracting it adds to the total: 337.72 is bigger than 184.96 + 90.25. An obtuse angle stretches the side opposite it, which makes this a useful check.
AC = $\sqrt{337.72…}$ = 18.377…

The rule gives the square of the side, so the square root is a line of its own. Only the positive root means anything — AC is a length.
AC = 18.4 cm (to 3 s.f.)

Round once, at the end, and state the units. AC is longer than either of the other two sides, as it must be: it faces the largest angle.

Walkthrough 10 — a side, then the area Intermediate

In triangle ABC, AB = 17 cm, AC = 12 cm and ∠BAC = 58°. Calculate (i) the length of BC and (ii) the area of triangle ABC.

Fig. 9.10

BC² = 17² + 12² − 2 × 17 × 12 × cos 58°

Two sides with the angle between them, and the third side wanted — the cosine rule's home ground. BC is opposite the 58°, so it is the side that stands alone.
= 289 + 144 − 408 × 0.52991… = 216.79…

This time cos 58° is positive, so the last term genuinely subtracts and 216.79 is less than 433. An acute angle pulls the opposite side in.
BC = $\sqrt{216.79…}$ = 14.723… = 14.7 cm (to 3 s.f.)

Square root, then round for the stated answer — but keep the full value on the calculator, because part (ii) may need it.
Area = ½ × 17 × 12 × sin 58°

Part (ii) does not use the answer to part (i) at all. The area formula wants two sides and the angle between them, and the question already gave exactly that — so going back to the original data avoids carrying any rounding error forward.
= 102 × 0.84804… = 86.500… = 86.5 cm² (to 3 s.f.)

Units squared for an area. Notice how much cleaner this was than finding a perpendicular height: with the included angle in hand, the height was never needed.

Finding an angle: three sides

When all three sides are known, the cosine rule is the only rule that will start, and it has a bonus the sine rule cannot match: the sign of the cosine settles the argument by itself. There is no ambiguous case here at all.

A negative cosine means an obtuse angle, and that is the end of it. Different angles between 0° and 180° have different cosines, so cos⁻¹ returns the one correct answer, acute or obtuse as the sign says.

Walkthrough 11 — the largest angle from three sides Basic

In triangle PQR, PQ = 7 cm, QR = 13 cm and PR = 9 cm. Find the size of the largest angle.

The largest angle is opposite the longest side, so it is ∠QPR, opposite QR = 13 cm.

Deciding which angle to compute is the whole of the first mark, and it takes no calculation: in any triangle the biggest angle faces the longest side. Say which angle you are finding before finding it.
cos ∠QPR = $\frac{7^{2} + 9^{2} - 13^{2}}{2 \times 7 \times 9}$

The rearranged cosine rule, with the square of the side opposite the wanted angle — 13² — as the one being subtracted, and the two sides that meet at P in the denominator.
= $\frac{49 + 81 - 169}{126}$ = $\frac{- 39}{126}$ = −0.30952…

The top is negative, because 13² is bigger than 7² + 9². By the converse of Pythagoras that already tells you the angle is obtuse, before the calculator is touched.
∠QPR = cos⁻¹(−0.30952…) = 108.0° (to 1 d.p.)

Inverse cosine returns the obtuse angle directly, so there is no second candidate to test and nothing to reject. Write the answer to 1 decimal place, keeping the ".0" — it shows the accuracy asked for.

Walkthrough 12 — a quadrilateral, cut into two triangles Advanced

In the quadrilateral ABCD, AB = 6.5 cm, BC = 9.2 cm, ∠ABC = 74°, ∠ACD = 41° and ∠ADC = 96°. Calculate (i) AC, (ii) AD and (iii) the area of the quadrilateral ABCD.

Fig. 9.11

AC² = 6.5² + 9.2² − 2 × 6.5 × 9.2 × cos 74° = 93.923…, so AC = 9.6914… = 9.69 cm (to 3 s.f.)

The diagonal AC is the bridge between the two triangles, so it is always the first thing to find. In triangle ABC the 74° is between the two known sides, which is the cosine rule's case.
In triangle ACD: ∠DAC = 180° − 96° − 41° = 43°

Angle sum, to complete the second triangle. With two angles known, triangle ACD now has a side (AC) paired with the angle facing it (96°), so the sine rule can take over.
$\frac{A D}{\sin 41°} = \frac{9.6914…}{\sin 96°}$ so AD = 6.3931… = 6.39 cm (to 3 s.f.)

AD is opposite the 41° and AC is opposite the 96°. The unrounded 9.6914… goes in, not 9.69 — that is the "shown to a higher degree of accuracy" rule doing real work.
Area of ABC = ½ × 6.5 × 9.2 × sin 74° = 28.741…

Back to the original data for the first half of the area, with 74° the included angle between AB and BC. No rounded value is involved anywhere in this line.
Area of ACD = ½ × 9.6914… × 6.3931… × sin 43° = 21.127…

The second half uses AC and AD with the 43° between them — the angle worked out two lines ago, which is exactly why it was worth finding. Both lengths go in unrounded.
Area of ABCD = 28.741… + 21.127… = 49.869… = 49.9 cm² (to 3 s.f.)

A quadrilateral has no area formula of its own here: the diagonal split it into two triangles, and the areas simply add. Round only now, and write cm².

Choosing between the two rules

The choice between the two rules is decided by what you have been given, and by nothing else. Read down the first column until the row matches your question.

You are given…	Is there a side with its opposite angle?	Use
two angles and one side	yes, or you can get one from the angle sum	Sine Rule
two sides and a non-included angle	yes	Sine Rule — then test the obtuse alternative
two sides and the included angle	no	Cosine Rule, to get the third side
all three sides	no	Cosine Rule, rearranged, to get an angle
two sides and the included angle, and the area is wanted	—	½`ab` sin `C` — neither rule is needed

One question can want both rules. Cosine rule for the missing third side, then sine rule for a second angle. Carry the unrounded value forward; a stored value written down keeps 5 significant figures, not 3.

After a cosine rule, prefer the cosine rule again. With all three sides now known, the rearranged cosine rule settles a second angle outright; the sine rule would put you back to testing a supplement.

More detail

Carrying the unrounded value forward is the paper's own instruction: an answer that has to be shown correct to a stated accuracy must be "first shown to a higher degree of accuracy". Using the rearranged cosine rule for the second angle avoids the ambiguous case altogether.

Check yourself

In triangle ABC, AB = 8 cm, BC = 11 cm and ∠ABC = 47°. Find AC.
Answer
The angle is between the two sides, so use the cosine rule: AC² = 8² + 11² − 2 × 8 × 11 × cos 47° = 64 + 121 − 120.03… = 64.968…, so AC = 8.06 cm (to 3 s.f.).
The sides of a triangle are 5 cm, 6 cm and 10 cm. Find its largest angle, and say what its sign tells you.
Answer
The largest angle faces the 10 cm side. cos θ = (5² + 6² − 10²) ÷ (2 × 5 × 6) = (−39) ÷ 60 = −0.65, so θ = 130.5° (to 1 d.p.). The negative cosine says at once that the angle is obtuse, so the triangle is obtuse-angled.
Which rule would you reach for in each case? (a) three sides, an angle wanted; (b) two angles and a side, a side wanted; (c) two sides and the angle between them, the area wanted.
Answer
(a) cosine rule rearranged — there is no side–angle pair to start a sine rule.
(b) sine rule — find the third angle first if the pairing needs it.
(c) neither: ½ab sin C gives the area straight from what you have.
In triangle PQR, PQ = 15 cm, QR = 8 cm and ∠PQR = 112°. Find PR and then the area.
Answer
PR² = 15² + 8² − 2 × 15 × 8 × cos 112° = 225 + 64 + 89.905… = 378.90…, so PR = 19.5 cm (to 3 s.f.). Area = ½ × 15 × 8 × sin 112° = 55.631… = 55.6 cm². Note the + in the first line: cos 112° is negative.
Explain why a triangle with sides 3 cm, 4 cm and 9 cm cannot exist, and what the cosine rule does if you try it anyway.
Answer
3 + 4 = 7, which is less than 9, so the two short sides cannot reach. The cosine rule gives cos θ = (9 + 16 − 81) ÷ 24 = −2.333…, and no angle has a cosine below −1, so the calculator refuses — which is the arithmetic telling you the same thing.

Past-paper questions

G3 N2019 · Paper 2 · Q2 8 marks

X is a point inside triangle ABC. AB = 7.5 cm, BC = 8.3 cm and BX = 6.4 cm. Angle AXB = 112° and angle XBC = 27°.

Calculate CX. [3]
Calculate angle XAB. [2]
Calculate the area of triangle ABC. [3]

Worked solution

(a) Triangle BXC has two sides, BX = 6.4 and BC = 8.3, with the 27° between them — the cosine rule:

${C X}^{2} = {6.4}^{2} + {8.3}^{2} - 2 \times 6.4 \times 8.3 \times \cos 27° = 15.1894…$

$C X = 3.8973…$ = 3.90 cm (to 3 s.f.).

(b) Triangle AXB has the 112° at X facing AB = 7.5, and BX = 6.4 faces the angle XAB — a side–angle pair each way, so the sine rule:

$\frac{\sin ∠ X A B}{6.4} = \frac{\sin 112°}{7.5}$ , so $\sin ∠ X A B = 0.79119…$

angle XAB = 52.3° (to 1 d.p.). The obtuse partner, 127.7°, is rejected: the triangle already has a 112° angle, and two obtuse angles cannot fit inside 180°.

(c) The area formula needs the angle at B in the big triangle, which is angle ABX plus the 27°:

angle ABX = 180° − 112° − 52.297…° = 15.702…°, so angle ABC = 42.702…°

area of triangle ABC $= \frac{1}{2} \times 7.5 \times 8.3 \times \sin 42.702…° = 21.108…$ = 21.1 cm² (to 3 s.f.).

Why this works. X cuts the figure into two triangles that want different rules, so each part starts by deciding which triangle you are in: two sides with the angle wedged between them is the cosine rule, an angle facing a known side is the sine rule. Part (c) needs the whole angle at B, which is the 15.702…° left over plus 27°. Add the two before rounding.

Angles of elevation and depression

Nothing new is being calculated here — the trigonometry is still the trigonometry of 9.1 to 9.3. What is new is translating a sentence into a triangle, and the two words that do most of that work are elevation and depression.

The angle of elevation of an object from a point is the angle between the horizontal at that point and the line of sight up to the object. The angle of depression is the angle between the horizontal and the line of sight down to it.

Fig. 9.12

An observer's eye is at A, and AD is an imaginary horizontal line from the eye to the wall. Looking up at the top B, the angle BAD between the horizontal and the line of sight is the angle of elevation. Looking down at the foot C, the angle CAD is the angle of depression.

Both are measured from the horizontal, not from the vertical. A 40° angle of elevation is 40° up from level, leaving 50° between the line of sight and the upright. An angle measured from the tower itself is not an angle of depression.

Fig. 9.13

The angle of elevation of T from P and the angle of depression of P from T are alternate angles between two parallel horizontals, so they are always equal.

Move a depression angle into the triangle first. It is drawn outside the triangle you want; copy it to the bottom with the reason "alt. ∠s" beside it, and the rest is ordinary right-angled trigonometry.

Walkthrough 13 — a height from an angle of elevation Basic

A vertical mast FT stands on level ground. From a point P on the ground, 24 m from the foot F, the angle of elevation of the top T is 37°. Find (i) the height of the mast and (ii) the distance PT.

In triangle PFT, ∠PFT = 90°.

The mast is vertical and the ground is level, so they meet at a right angle — and that is what makes the whole question ordinary SOH-CAH-TOA. Naming the right angle first is the habit that will save you in 9.5.
tan 37° = $\frac{F T}{24}$

Relative to the 37°, the mast FT is the opposite side and the 24 m is the adjacent side, so tangent is the ratio that connects exactly those two. Choosing the ratio that avoids the hypotenuse keeps part (i) independent of part (ii).
FT = 24 tan 37° = 18.085… = 18.1 m (to 3 s.f.)

One multiplication, then round. A useful check: 37° is less than 45°, so the mast must be shorter than the 24 m distance — 18.1 m is.
cos 37° = $\frac{24}{P T}$ so PT = $\frac{24}{\cos 37°}$

PT is the hypotenuse and the 24 m is adjacent, so cosine is the pairing. Going back to the given 24 m rather than to the rounded 18.1 m keeps this part free of any error carried in.
PT = 30.051… = 30.1 m (to 3 s.f.)

The hypotenuse is the longest side, so it must exceed both 24 m and 18.1 m — and it does. Round once, at the end, and give the metres.

Walkthrough 14 — two angles of depression Intermediate

A vertical cliff LT is 84 m high, with L at sea level. From the top T, the angle of depression of a boat A is 51° and the angle of depression of a further boat B is 29°. The two boats and the foot of the cliff are in a straight line on the same side of the cliff. Find the distance AB.

Fig. 9.14

∠TAL = 51° and ∠TBL = 29° (alt. ∠s, horizontal at T parallel to the sea)

The depressions are marked at the top of the cliff, outside both triangles. Moving each of them down to the boat, with the parallel-lines reason stated, puts them inside right-angled triangles where a tangent can be used.
In triangle TLA: tan 51° = $\frac{84}{L A}$ so LA = $\frac{84}{\tan 51°}$ = 68.021…

The cliff is opposite the 51° and LA is adjacent to it, so tangent again. LA is not asked for, so it is kept unrounded on the calculator rather than written to 3 s.f.
In triangle TLB: LB = $\frac{84}{\tan 29°}$ = 151.54…

Exactly the same line for the second boat. The smaller angle of depression belongs to the boat that is further away, and indeed 151.5 > 68.0.
AB = LB − LA = 151.54… − 68.021… = 83.518… = 83.5 m (to 3 s.f.)

Both boats are on the same side of the cliff, so the gap between them is a difference of the two distances, not a sum. Subtract the unrounded values and round once, here at the end.

Where this comes from

The two angles are tied to each other. The horizontal at the top of a tower and the horizontal at the ground are parallel, so a pair of alternate angles appears wherever a sloping line of sight crosses them — which is why the depression angle can be moved down to the foot of the triangle and used there.

Do that move once, at the start, and everything after it is the trigonometry of 9.1 to 9.3 with nothing new in it.

The sketch decides one more thing. Two objects on the same side of a tower or cliff give a difference of distances; on opposite sides the same line would be an addition.

Past-paper questions

G3 N2015 · Paper 2 · Q9 13 marks

The diagram shows a field ABCD on horizontal ground, crossed by a path BD. AB = 95 m, AD = 60 m and BC = CD. Angle BAD = 90° and angle BCD = 40°.

Show that angle ADC = 127.7°, correct to one decimal place. [2]
Find CD. [4]
The land is valued at $40 000 per hectare. Given that 1 hectare = 10 000 square metres, calculate the value of the field. [4]
A bird is hovering vertically above B. The angle of elevation of the bird from A is 18°. Find the angle of elevation of the bird from D. [3]

Worked solution

(a) In right-angled triangle ABD, $\tan ∠ A D B = \frac{95}{60}$ , so angle ADB = 57.724…°.

Triangle BCD is isosceles (BC = CD) with apex 40°, so angle BDC = (180° − 40°) ÷ 2 = 70° (base angles).

angle ADC = 57.724…° + 70° = 127.724…° = 127.7° (shown).

(b) By Pythagoras, ${B D}^{2} = 60^{2} + 95^{2} = 12 625$ , so BD = 112.361… m.

In triangle BCD, CD faces the 70° at B and BD faces the 40° at C, so by the sine rule

$C D = \frac{112.361… \times \sin 70°}{\sin 40°} = 164.26…$ = 164 m (to 3 s.f.).

(c) Split the field along the path:

area of triangle ABD $= \frac{1}{2} \times 95 \times 60 = 2850$ m²

area of triangle BCD $= \frac{1}{2} \times {(164.26…)}^{2} \times \sin 40° = 8671.72…$ m²

total area = 11 521.72… m² = 1.152172… hectares, so the value is 1.152172… × $40 000 = $46 086.90… = $46 100 (to 3 s.f.).

(d) Let the bird be at T, vertically above B. From A the horizontal distance is AB = 95, so

$B T = 95 \tan 18° = 30.867…$ m

From D the horizontal distance is BD = 112.361… m, so the angle of elevation is

tan (angle of elevation from D) $= \frac{30.867…}{112.361…}$ , giving 15.361…° = 15.4° (to 1 d.p.).

Why this works. The path BD is the hinge: it splits the field into a right-angled triangle and an isosceles one, and it is shared, so store 112.361… and reuse it in (b) and (d). In (a) angle ADC is the sum of the two angles at D, which is what "show that" wants. In (c) one hectare is 10 000 m².

G3 N2020 · Paper 1 · Q18 4 marks

The diagram represents a tower, AB, built on horizontal ground. The height of the tower is 45 m. From a point C, the angle of elevation of the top of the tower is 36°. Point D is 22 m from C and BCD is a straight line.

Calculate the angle of elevation of the top of the tower from the point D. [4]

Worked solution

In right-angled triangle ABC, 45 is opposite the 36° and BC is adjacent to it:

$B C = \frac{45}{\tan 36°} = 61.937…$ m

BCD is a straight line, so BD = 61.937… + 22 = 83.937… m, and in right-angled triangle ABD

tan (angle of elevation from D) $= \frac{45}{83.937…}$ , giving 28.196…° = 28.2° (to 1 d.p.).

Why this works. Both elevations stand on the same 45 m, so only the horizontal distance changes and finding it ends the question. The four marks pay for three decisions: tan (opposite and adjacent), dividing by tan 36° because the unknown is on the bottom, and adding the 22 m. D is further out, so the answer comes out below 36°; keep 61.937… unrounded before adding.

Bearings

A bearing is how a question states a direction. The convention has three parts, and the marks in a bearings question sit on those rather than on the trigonometry.

The bearing of a point B from a point A is the angle measured (1) from the north line at A, (2) clockwise, (3) to the line AB. It is always written as a three-digit number when it is a whole number of degrees — so a turn of 42° is written 042°, not 42°.

Fig. 9.15

Only the small angle to the nearer of the north and south lines is marked, as it usually is in an exam diagram; the bearing is then one subtraction away. Reading clockwise from north: A is on a bearing of 062°; B is on 180° − 37° = 143°; C is on 180° + 53° = 233°; and D is on 360° − 50° = 310°.

The four quarters. North is 000° (or 360°), east 090°, south 180° and west 270°. A bearing runs clockwise all the way round to just under 360°, so it is never negative and never anticlockwise.

Three digits. Writing 3° as 003° keeps it from reading as 30° or 300°. A bearing that is not a whole number keeps its decimal: 183.1° is written just like that.

Because the north lines at two different points are parallel, the bearing of A from B is never the same as the bearing of B from A. The two differ by exactly a half-turn.

Fig. 9.16

If the bearing of B from A is 070°, then the bearing of A from B is 070° + 180° = 250°. Adding 180° would take a bearing past 360°, so in that case subtract 180° instead.

"The bearing of X from Y" — stand at Y. The second point named carries the north line; the first is what you are looking at. Start every bearings solution by drawing that north line.

Walkthrough 15 — reading bearings off a diagram Basic

In the diagram, A lies 63° east of the north line at O and B lies 48° west of it. Find the bearing of (i) A from O, (ii) B from O, (iii) O from A and (iv) O from B.

Fig. 9.17

Bearing of A from O = 063°

Measuring clockwise from the north line at O reaches OA after 63°, so the angle marked is the bearing. Three digits, so the leading zero is written.
Bearing of B from O = 360° − 48° = 312°

B is to the west of north, so going clockwise from north has to travel almost all the way round before it arrives. The 48° marked is measured anticlockwise, and a bearing never is — hence the subtraction from a full turn.
Bearing of O from A = 063° + 180° = 243°

Now you are standing at A, and the north line at A is parallel to the one at O. Reversing the direction of travel turns the bearing through a half-turn; adding is right here because 63 + 180 is still under 360.
Bearing of O from B = 312° − 180° = 132°

The same half-turn, the other way: 312 + 180 = 492 is past a full turn and so is not a bearing, so 180° is subtracted instead. Check it: O is south-east of B, and 132° lies between 090° and 180°.

More detail

Reading "the bearing of X from Y" the wrong way round gives an answer exactly 180° out, because the two north lines are parallel and the direction has simply been reversed. Drawing the north line at the point you are standing on settles it before any arithmetic starts.

The convention's three parts — from north, clockwise, three digits — are where most bearings marks are decided; the trigonometry underneath is the ordinary trigonometry of 9.2 and 9.3.

Past-paper questions

G3 N2024 · Paper 1 · Q10 5 marks

A, B and C are three points on horizontal ground. Angle ABC = 58°, angle BAC = 90° and BC = 36 m.

Calculate the distance AC. [2]
Calculate the perimeter of triangle ABC. [2]
The point D is on a bearing of 116° from B. Find the bearing of B from D. [1]

Worked solution

(a) The right angle is at A, so BC = 36 is the hypotenuse and AC is opposite the 58°:

$A C = 36 \sin 58° = 30.529…$ = 30.5 m (to 3 s.f.).

(b) AB is adjacent to the 58°:

$A B = 36 \cos 58° = 19.077…$ m

perimeter = 36 + 30.529… + 19.077… = 85.606… = 85.6 m (to 3 s.f.).

(c) The two north lines are parallel, so the return direction turns through a straight angle: bearing of B from D = 116° + 180° = 296°.

Why this works. Parts (a) and (b) are marked on the ratio you choose, so name the hypotenuse first: it is the side facing the right angle, here BC. Then AC is opposite the 58° (sine) and AB is adjacent (cosine); Pythagoras is equally acceptable, given the unrounded 30.529…. Part (c) is one mark for a back bearing: 116 + 180 = 296, still under 360.

G3 N2022 · Paper 1 · Q9 4 marks

In this scale drawing, A and B are two ports.

A boat leaves A and sails on a bearing of 040° to a lighthouse 110 km away.

Mark and label on the drawing the position, L, of the lighthouse. [2]
A second boat leaves B to meet the first boat at sea. When they meet, the second boat has sailed the shortest possible distance. Find the bearing the second boat sails on. [2]

Worked solution

(a) The scale is 1 cm to 10 km, so 110 km is drawn as 11 cm. Put the protractor on the north line at A, measure 40° clockwise from north, and mark L 11 cm from A along the 040° line.

(b) The first boat's whole voyage lies on the line AL, so the second boat meets it somewhere on that line. The shortest route from a point to a line is the perpendicular to it, so the second boat sails from B at right angles to AL.

Turning 90° from the 040° direction gives 040° + 90° = 130° or 040° − 90° = −50°, that is 310°. The drawing shows B lying to the east of the line AL, so the boat must head back towards the west of north:

bearing = 310°.

Why this works. "Shortest possible distance" is the instruction to use a perpendicular, and a right-angle turn changes a bearing by 90°, so the answer is 040 ± 90. Both marks hang on the sign, and the picture settles it: B is east of the track, so the boat sails north-west. In (a), convert before drawing — 110 km ÷ 10 km per cm = 11 cm.

Bearings problems — the triangle inside the compass

A bearings question is a triangle question with a compass wrapped round it. Unwrapping it is a fixed routine, worth doing on paper every time.

1. Sketch the points and draw a north line at every point a bearing is measured from. 2. Mark each given bearing as a clockwise angle from its own north line. 3. Use parallel-line angle facts — alternate angles, co-interior angles adding to 180°, angles on a straight line — to turn those bearings into the interior angles of the triangle. 4. Only now choose the sine or cosine rule.

Shortcut 1: B due north of A. The line AB is the north line at both points, so the bearing of any third point from A is already the interior angle at A.

Shortcut 2: the same pair, seen from B. The direction back to A is due south, so the interior angle at B is 180° − (the bearing from B) whenever that bearing points east of the line.

Walkthrough 16 — bearings into a sine rule Intermediate

Three points A, B and C lie on level ground. B is due north of A, the bearing of C from A is 058° and the bearing of C from B is 137°. Given that AB = 240 m, find (i) ∠ACB and (ii) the distance AC.

∠BAC = 58°

Because B is due north of A, the north line at A runs along AB itself — so the 058° bearing, measured from that line to AC, is already the interior angle of the triangle at A. Nothing has to be converted.
∠ABC = 180° − 137° = 43°

Standing at B, A is due south, which is a bearing of 180°. The bearing to C is 137°, so swinging from BC round to BA covers the remaining 43° — angles on a straight line.
∠ACB = 180° − 58° − 43° = 79°

Angle sum of the triangle, now that both other angles are known. This also completes a side–angle pair for the next step: 79° sits opposite the one side you were given.
$\frac{A C}{\sin 43°} = \frac{240}{\sin 79°}$

Two angles and a side, so the sine rule. AC faces the 43° at B and the 240 m faces the 79° at C — each fraction holds a side over the sine of the angle opposite it.
AC = $\frac{240 \sin 43°}{\sin 79°}$ = 166.74… = 167 m (to 3 s.f.)

AC faces a smaller angle than AB does, so it should come out shorter than 240 m — and it does. Three significant figures here means no decimal point at all.

Walkthrough 17 — two ships, and a bearing as the answer Advanced

Two ships leave a port P at the same time. Ship A sails on a bearing of 042° at 18 km/h; ship B sails on a bearing of 155° at 24 km/h. After 2.5 hours, find (i) the distance between the two ships and (ii) the bearing of B from A.

Fig. 9.18

PA = 18 × 2.5 = 45 km and PB = 24 × 2.5 = 60 km

The speeds are not lengths, so the triangle does not exist until distance = speed × time has been applied to both. Both ships have travelled for the same 2.5 hours, which is what makes a single triangle possible.
∠APB = 155° − 42° = 113°

Both bearings are measured clockwise from the same north line at P, so the angle between the two courses is simply the difference of the bearings. This is the only place in the question where the compass is needed for the first part.
AB² = 45² + 60² − 2 × 45 × 60 × cos 113° = 7734.9…, so AB = 87.948… = 87.9 km (to 3 s.f.)

Two sides with the angle between them and no side–angle pair anywhere — the cosine rule. cos 113° is negative, so the last term adds and AB comes out longer than either course, as an obtuse angle demands.
$\frac{\sin ∠ P A B}{60} = \frac{\sin 113°}{87.948…}$ so ∠PAB = 38.901…°

A bearing is an angle, so an angle inside the triangle has to come first. The unrounded 87.948… is used. There is no ambiguity to test: ∠APB is already obtuse, so no other angle in this triangle can be.
The angle between AP and the south direction at A is 42° (alt. ∠s, north lines parallel).

The north line at A is parallel to the one at P, so the 042° at P reappears at A as the angle between AP and due south. That is the bridge from an angle inside the triangle back out to a compass direction.
Bearing of B from A = 180° + (42° − 38.901…°) = 183.1° (to 1 d.p.)

From A, AB lies 38.9° from AP on the eastern side, so it is 42 − 38.9 = 3.1° west of due south, and bearings just west of south are 180° plus that. Check the sketch: 183.1° is almost due south.

Check yourself

Write down the bearing of a point that lies due west, and the bearing of a point 7° east of north.
Answer
Due west is 270°. Seven degrees clockwise from north is 007° — the two leading zeros are part of the answer.
The bearing of Q from P is 218°. Find the bearing of P from Q.
Answer
218 + 180 = 398, which is more than 360, so subtract instead: 218° − 180° = 038°. (Check: Q is south-west of P, so P is north-east of Q, and 038° is north-east.)
From a point 40 m from the foot of a tower on level ground, the angle of elevation of the top is 54°. Find the height of the tower.
Answer
height = 40 tan 54° = 55.055… = 55.1 m (to 3 s.f.).
From the top of a 60 m lighthouse the angle of depression of a boat is 23°. How far is the boat from the foot of the lighthouse?
Answer
The angle of elevation of the top from the boat is also 23° (alt. ∠s), so distance = 60 ÷ tan 23° = 141.35… = 141 m (to 3 s.f.).
Three points X, Y and Z are on level ground. Y is due north of X, the bearing of Z from X is 071° and the bearing of Z from Y is 118°. Find ∠XZY.
Answer
∠ZXY = 71° (the north line at X is XY), and ∠XYZ = 180° − 118° = 62° (angles on a straight line, since X is due south of Y). So ∠XZY = 180° − 71° − 62° = 47°.
A walker leaves camp on a bearing of 130° and walks 6 km, then turns and walks 4 km on a bearing of 220°. How far is she from camp?
Answer
The two bearings differ by 90°, and at the turning point the direction back to camp is 130 + 180 = 310°, so the angle between the return line and the new course is 310 − 220 = 90° — a right angle. So Pythagoras is enough: distance = $\sqrt{6^{2} + 4^{2}}$ = $\sqrt{52}$ = 7.2111… = 7.21 km (to 3 s.f.).

Past-paper questions

G3 N2015 · Paper 1 · Q18 4 marks

The diagram shows the positions of three towns P, Q and R. PQ is 46.5 km, PR is 23.6 km and QR is 56.0 km. The bearing of Q from P is 041°.

Calculate the bearing of R from P. [4]

Worked solution

All three sides of triangle PQR are known, so the cosine rule gives the angle at P:

$\cos ∠ Q P R = \frac{{46.5}^{2} + {23.6}^{2} - {56.0}^{2}}{2 \times 46.5 \times 23.6} = −0.18989…$

angle QPR = 100.946…°

R lies clockwise from PQ, so the bearing of R from P is measured on from the 041°:

bearing = 041° + 100.946…° = 141.946…° = 141.9° (to 1 d.p.).

Why this works. Three sides and no angles can start only one way: the cosine rule, rearranged for the angle. The negative cosine is not an error — it says angle QPR is obtuse, which the diagram agrees with. The last mark is not trigonometry at all: the 041° carries north round to PQ and the 100.9° carries on in the same direction, so the two add.

G3 N2017 · Paper 2 · Q5 10 marks

Points A, B and C are at sea level. A is 104 m from C on a bearing of 064°. B is 135 m from C on a bearing of 122°.

Calculate AB. [3]
Calculate the bearing of A from B. [3]
A boat, T, travels in a straight line from C to B. Calculate the shortest distance of T from A during this journey. [2]
C is a point at the base of a vertical cliff. D is the point on the top of the cliff vertically above C. The angle of depression of B from D is 27°. Calculate the height of the cliff. [2]

Worked solution

(a) Both bearings are measured from the same north line at C, so the angle between the two directions is their difference:

angle ACB = 122° − 64° = 58°

Two sides with the angle between them — the cosine rule:

${A B}^{2} = 104^{2} + 135^{2} - 2 \times 104 \times 135 \times \cos 58° = 14 160.86…$

$A B = 118.999…$ = 119 m (to 3 s.f.).

(b) The sine rule gives the angle at B:

$\frac{\sin ∠ A B C}{104} = \frac{\sin 58°}{118.999…}$ , so angle ABC = 47.829…°

The north line at B is parallel to the one at C, and the bearing of C from B is the back bearing 122° + 180° = 302°. Swinging on a further 47.829…° reaches A:

bearing of A from B = 302° + 47.829…° = 349.829…° = 349.8° (to 1 d.p.).

(c) The shortest distance from A to the line CB is the perpendicular. In the right-angled triangle it makes with CA = 104 and the 58° at C,

shortest distance $= 104 \sin 58° = 88.197…$ = 88.2 m (to 3 s.f.).

(d) The angle of depression of B from D equals the angle of elevation of D from B (alternate angles between the horizontal at D and the ground). In right-angled triangle DCB, with CB = 135 m on the ground,

height $= 135 \tan 27° = 68.785…$ = 68.8 m (to 3 s.f.).

Why this works. Both bearings start from the same north line at C, so the included angle is a subtraction, and it unlocks (a) and (c). In (b) the direction of travel reverses, so build the back bearing at B first; the parallel north lines license it. In (c) "shortest distance" means the perpendicular; in (d) the depression transfers to an equal elevation.

G3 N2018 · Paper 1 · Q24 4 marks

The diagram shows the positions of three cities, Singapore (S), New Delhi (D) and Manila (M). SM = 2392 km, SD = 4140 km and angle SDM = 29°.

Calculate acute angle DMS. [2]
The bearing of New Delhi from Manila is 290°. Find the bearing of New Delhi from Singapore. [2]

Worked solution

(a) SM = 2392 faces the 29° at D, and SD = 4140 faces the angle at M — a side–angle pair each way, so the sine rule:

$\frac{\sin ∠ D M S}{4140} = \frac{\sin 29°}{2392}$ , so $\sin ∠ D M S = 0.839093…$

acute angle DMS = 57.044…° = 57.0° (to 1 d.p.).

(b) A bearing of 290° is 70° short of a full turn, so the angle at M between its north line and MD, measured anticlockwise, is 360° − 290° = 70°.

The north lines at D and M are parallel, so the bearing of M from D is 180° − 70° = 110° (co-interior angles).

S lies a further 29° clockwise from DM, so the bearing of S from D is 110° + 29° = 139°, and the return journey is the back bearing:

bearing of New Delhi from Singapore = 139° + 180° = 319°.

Why this works. Part (a) is the sine rule's signature, two sides and a non-included angle, and the word acute closes the obtuse branch, so no rejection argument is needed. Part (b) is pure parallel-line work: 290° is easier used as the 70° on the other side of north, carried along DM as co-interior angles. The 57.0° from (a) never enters it.

G3 N2019 · Paper 1 · Q14 3 marks

The diagram shows the positions of three cities, Singapore (S), Delhi (D) and Tokyo (T). Angle SDT = 55°. The bearing of Delhi from Singapore is 317° and the bearing of Tokyo from Singapore is 044°.

Find the bearing of Singapore from Delhi. [1]
Find the bearing of Delhi from Tokyo. [2]

Worked solution

(a) The bearing of Delhi from Singapore is 317°, which is more than 180°, so the back bearing is found by subtracting:

bearing of Singapore from Delhi = 317° − 180° = 137°.

(b) Both bearings at S are measured from the same north line, and the angle between them going clockwise from ST round to SD is 317° − 44° = 273°, which is reflex. The angle inside the triangle is therefore

angle DST = 360° − 273° = 87°

angle DTS = 180° − 55° − 87° = 38° (angle sum of a triangle)

At T, the bearing of S is the back bearing of 044°, that is 044° + 180° = 224°, and D lies a further 38° clockwise:

bearing of Delhi from Tokyo = 224° + 38° = 262°.

Why this works. The whole thing turns on one check: is the difference of the bearings the angle inside the triangle, or its reflex? Here 317 − 44 = 273, which no triangle can hold, so the interior angle is the rest of the full turn. Drawing the north line at S inside the triangle makes that visible; the rest is the angle sum and one back bearing.

G3 N2020 · Paper 2 · Q4 9 marks

ABCD is a park on horizontal ground crossed by a path AC. A is due north of D. AB = 660 m, BC = 950 m and AD = 480 m. Angle ABC = 80°, angle ACD = 24° and angle ADC is obtuse.

Show that AC = 1060 m, correct to 3 significant figures. [3]
Calculate the bearing of D from C. [3]
Sita goes for a jog in the park at a speed of 9.5 km/h. She starts at A, jogs around the edge to B then C, and then directly back to A. Calculate the time it takes her to jog this route. Give your answer in minutes and seconds, correct to the nearest ten seconds. [3]

Worked solution

(a) In triangle ABC, the 80° is between the two known sides, so the cosine rule:

${A C}^{2} = 660^{2} + 950^{2} - 2 \times 660 \times 950 \times \cos 80°$

$A C = 1058.46…$ = 1060 m (shown).

(b) In triangle ACD, AC faces the angle at D and AD = 480 faces the 24°, so

$\frac{\sin ∠ A D C}{1058.46…} = \frac{\sin 24°}{480}$ , so $\sin ∠ A D C = 0.89691…$

The question says angle ADC is obtuse, so take the obtuse root:

angle ADC = 180° − 63.754…° = 116.245…°

A is due north of D, so DA is the north line at D, and the bearing of C from D is the angle ADC itself. The bearing of D from C is its back bearing:

bearing = 116.245…° + 180° = 296.245…° = 296.2° (to 1 d.p.).

(c) The route is AB + BC + CA:

distance = 660 + 950 + 1058.46… = 2668.46… m = 2.66846… km

$time = \frac{2.66846…}{9.5} = 0.28089…$ h = 1011.2… s = 16 min 51.2… s

to the nearest ten seconds, 16 minutes 50 seconds.

Why this works. AC is the hinge, so store it unrounded — the printed 1060 turns the bearing into 296.1°. In (b) the sine rule offers two angles and the question says which to take, and "A is due north of D" puts the north line along DA. In (c), convert to kilometres before dividing, and read the rounding literally: 51.2 s goes down to 50.

G3 N2023 · Paper 1 · Q23 7 marks

A, B and C are three points on horizontal ground. BA = 63 m and the bearing of A from B is 060°. BC = 95 m and angle ABC = 64°.

Find the bearing of B from C. [1]
Calculate the distance AC. [3]
Calculate the bearing of C from A. [3]

Worked solution

(a) Both angles at B are measured from the same north line, so the bearing of C from B is 060° + 64° = 124°. The bearing of B from C is the back bearing:

124° + 180° = 304°.

(b) Two sides with the 64° between them — the cosine rule:

${A C}^{2} = 63^{2} + 95^{2} - 2 \times 63 \times 95 \times \cos 64° = 7746.69…$

$A C = 88.015…$ = 88.0 m (to 3 s.f.).

(c) The sine rule gives the angle at A:

$\frac{\sin ∠ B A C}{95} = \frac{\sin 64°}{88.015…}$ , so angle BAC = 75.958…°

The bearing of B from A is 060° + 180° = 240°, and C lies 75.958…° anticlockwise from AB, so

bearing of C from A = 240° − 75.958…° = 164.041…° = 164.0° (to 1 d.p.).

Why this works. Every part uses the same north line: in (a) the two angles at B stack; in (b) the 64° is already the included angle. Part (c) turns an angle inside the triangle into a bearing against the back bearing 240° of AB, and the picture decides whether to add or subtract. The obtuse partner goes because cos ∠BAC = +0.2426 is positive.

Reducing three dimensions to two

There is no three-dimensional trigonometry in this syllabus: every question is solved by finding a flat triangle inside the solid and doing ordinary two-dimensional work in it. The difficulty is entirely in the finding.

The basic technique for a three-dimensional problem is to reduce it to a problem in a plane: identify the right-angled triangle that contains the side or the angle you have been asked for, redraw that triangle on its own, and work in the redrawn version.

A right angle in a 3-D drawing rarely looks like one. Flat paper distorts angles: a 90° corner can be drawn as 60° or as 120° and still be a genuine right angle. Decide from the solid, using one of the two tests below.

Test 1: find a rectangle. If the three points all lie on one rectangular face, any angle of the triangle sitting at a corner of that rectangle is 90°. A diagonal of a rectangle makes a right-angled triangle with two of its sides.

Test 2: find a normal. A normal to a plane is perpendicular to every line in it. A vertical edge or mast is a normal to the horizontal base, so it meets every line on that base, diagonals included, at 90°.

Fig. 9.19

In the cuboid ABCDEFGH, the edge CG is vertical and AC is a diagonal of the horizontal base, so CG is a normal to that base and ∠ACG = 90° — even though the drawing makes it look much bigger. Triangle ACG is right-angled, and it is the triangle that contains the space diagonal AG.

The two-stage pattern. Stage 1: Pythagoras in the horizontal base, for a diagonal you cannot measure directly. Stage 2: that diagonal is the base of a vertical right-angled triangle, and a ratio finishes it. Stage 1 produces a length nobody asked for.

Walkthrough 18 — a pyramid with a vertical edge Basic

The figure shows a pyramid with a rectangular base PQRS and vertex V vertically above R. Given that PQ = 15 cm, QR = 8 cm and VR = 6 cm, find (i) ∠VQR and (ii) ∠VPR.

Fig. 9.20

In triangle VQR, ∠VRQ = 90°.

VR is vertical, so it is a normal to the horizontal base — and a normal is perpendicular to every line in the base, including RQ. That is why the angle is a right angle, and the reason is worth writing.
tan ∠VQR = $\frac{V R}{Q R}$ = $\frac{6}{8}$ so ∠VQR = 36.9° (to 1 d.p.)

In the redrawn flat triangle, VR is opposite the angle wanted and QR is adjacent, so tangent connects them. Nothing outside triangle VQR is involved at all.
In triangle PQR, ∠PQR = 90°, so PR² = 15² + 8² = 225 + 64 = 289, so PR = 17 cm.

Part (ii) needs triangle VPR, and one of its sides is PR, a diagonal of the base that was not given. The base is a rectangle, so the angle at Q is 90° and Pythagoras supplies it.
In triangle VPR, ∠VRP = 90°.

The same normal argument as before, now applied to RP. RP is a line in the horizontal base, and VR is perpendicular to every such line — which is exactly why the base diagonal was worth finding.
tan ∠VPR = $\frac{V R}{P R}$ = $\frac{6}{17}$ so ∠VPR = 19.4° (to 1 d.p.)

Stage 2: the base diagonal is the adjacent side of a vertical right-angled triangle. The angle is smaller than the 36.9° of part (i), as it must be — the same 6 cm rise is being spread over a longer horizontal run.

More detail

Almost every cuboid or pyramid question runs on that two-stage pattern, and the length produced by stage 1 is needed even though the question never mentions it.

The reason the two tests matter is that a solid drawn on flat paper distorts every angle that is not in the plane of the page, so a triangle that looks right-angled may not be one, and one that looks nothing like it may be. The rectangle test and the normal test decide the question from the solid itself.

Past-paper questions

G3 N2018 · Paper 2 · Q8(a), (b) 6 marks

The diagram shows a pyramid ABCDE. The base of the pyramid is a square of side 15 cm. E is vertically above the centre of the square base. The vertical height of the pyramid is 20 cm.

Show that AE = 22.6 cm, correct to three significant figures. [3]
Calculate angle BAE. [3]

Worked solution

(a) Let M be the centre of the base. AM is half of the diagonal of the square, and the diagonal is

$\sqrt{15^{2} + 15^{2}} = 21.213…$ cm, so AM = 10.606… cm.

EM is vertical, so triangle AME is right-angled at M:

${A E}^{2} = {(10.606…)}^{2} + 20^{2} = 512.5, A E = \sqrt{512.5} = 22.638…$ = 22.6 cm (shown).

(b) Triangle ABE is isosceles: AE = BE = 22.638… cm and AB = 15 cm. Drop the perpendicular from E to the midpoint N of AB, so AN = 7.5 cm and

$\cos ∠ B A E = \frac{7.5}{22.638…} = 0.33129…$

angle BAE = 70.652…° = 70.7° (to 1 d.p.).

Why this works. Every three-dimensional part is a hunt for a right-angled triangle lying in one flat plane. In (a) that plane is AME, and the work is seeing that AM is half the diagonal, not half the side. In (b) the face ABE is isosceles, so the perpendicular from E lands on the midpoint of AB; carry $\sqrt{512.5}$ forward, since the rounded 22.6 gives 70.6°.

G3 N2022 · Paper 2 · Q7 10 marks

The diagram shows a cuboid ABCDEFGH. AB = 24 cm, BC = 13 cm and AE = 8 cm. X is a point on EF such that EX : XF = 3 : 5.

Calculate the area of trapezium ABXE. [3]
Calculate XG. [2]
A pyramid is cut from the cuboid as shown. The base of the pyramid is triangle BGX. Calculate the area of triangle BGX. [5]

Worked solution

(a) EX : XF = 3 : 5 and EF = AB = 24 cm, so $E X = \frac{3}{8} \times 24 = 9$ cm.

ABXE is a trapezium with parallel sides AB = 24 and EX = 9, and perpendicular height AE = 8:

area $= \frac{1}{2} \times (24 + 9) \times 8$ = 132 cm².

(b) XF = 15 cm and FG = BC = 13 cm, and XFG is a right angle in the top face, so

$X G = \sqrt{15^{2} + 13^{2}} = \sqrt{394} = 19.849…$ = 19.8 cm (to 3 s.f.).

(c) Find the three sides of triangle BGX, each inside a flat face:

$B X = \sqrt{15^{2} + 8^{2}} = \sqrt{289} = 17$ cm (in the face ABFE),

$B G = \sqrt{13^{2} + 8^{2}} = \sqrt{233} = 15.264…$ cm (in the face BCGF), and $X G = \sqrt{394}$ cm from (b).

Three sides, no angle — the cosine rule, taken at G:

$\cos ∠ X G B = \frac{394 + 233 - 289}{2 \sqrt{394} \sqrt{233}} = 0.55777…$ , so angle XGB = 56.097…°

area $= \frac{1}{2} \times \sqrt{394} \times \sqrt{233} \times \sin 56.097…° = 125.73…$ = 126 cm² (to 3 s.f.).

Why this works. Triangle BGX lies in no face of the cuboid, so none of its angles can be assumed; but each of its three sides does lie in a flat face, so each is an ordinary Pythagoras calculation. Then it is an ordinary triangle: cosine rule, then ½ab sin C, with $\sqrt{394}$ and $\sqrt{233}$ left in surd form. In BX the horizontal step is XF = 15.

G3 N2023 · Paper 1 · Q26 4 marks

The diagram shows an open cuboid measuring 12 cm by 8 cm by 6 cm. A rod is placed inside the cuboid against the face AEHD. In this position, 7 cm of the rod is outside the cuboid. The position of the rod is changed so that the length outside the cuboid is as short as possible.

Calculate this shortest length. [4]

Worked solution

Against the face AEHD the rod runs along that face's diagonal AH. The face measures 8 cm by 6 cm, so

$A H = \sqrt{8^{2} + 6^{2}} = 10$ cm

length of the rod = 10 + 7 = 17 cm

The longest straight line that fits inside the cuboid is the space diagonal AG. Working in two steps, ${A F}^{2} = 12^{2} + 8^{2}$ along the base, and then

$A G = \sqrt{12^{2} + 8^{2} + 6^{2}} = \sqrt{244} = 15.620…$ cm

shortest length outside = 17 − 15.620… = 1.379… = 1.38 cm (to 3 s.f.).

Why this works. Nothing can be done until the rod is measured: against the face AEHD the part inside is that face's diagonal, 8 and 6 giving 10, so the rod is 17 cm. The rest is the space diagonal $\sqrt{l^{2} + b^{2} + h^{2}}$ — Pythagoras twice. The largest face diagonal, $\sqrt{12^{2} + 8^{2}} = 14.42$ , leaves 2.58 cm, which is longer.

G3 N2024 · Paper 2 · Q5 9 marks

Points A, B, C and D are on horizontal ground. AB = 175 m, AC = 280 m, BC = 405 m and angle ACD = 32°.

B is on a bearing of 048° from A. D is due west of A.

Show that angle BAC = 124.1°, correct to one decimal place. [3]
Calculate AD. [3]
Point A is the base of a vertical mast AT. The angle of elevation of T from C is 34°. Calculate the angle of elevation of T from B. [3]

Worked solution

(a) Three sides are known, so the cosine rule, rearranged for the angle at A:

$\cos ∠ B A C = \frac{175^{2} + 280^{2} - 405^{2}}{2 \times 175 \times 280} = \frac{−55 000}{98 000}$

angle BAC = 124.140…° = 124.1° (shown).

(b) Go all the way round the point A. From the west direction AD clockwise to north is 90°, from north to AB is 48°, and from AB to AC is 124.140…°, so

angle DAC = 360° − 90° − 48° − 124.140…° = 97.859…°

angle ADC = 180° − 97.859…° − 32° = 50.140…°

By the sine rule in triangle ACD,

$A D = \frac{280 \sin 32°}{\sin 50.140…°} = 193.29…$ = 193 m (to 3 s.f.).

(c) The mast is vertical at A, so triangle TAC is right-angled at A with AC = 280 on the ground:

$A T = 280 \tan 34° = 188.86…$ m

Triangle TAB is right-angled at A too, with AB = 175, so the elevation of T from B satisfies

tan (angle) $= \frac{188.86…}{175}$ , giving 47.181…° = 47.2° (to 1 d.p.).

Why this works. Part (b) is the one to study: no triangle holds both D and the bearing, so the link is built at A from angles at a point totalling 360°. The sine rule then pairs AD with the 32° and the known 280 with the angle at D. In (c) both sight lines start from the mast at A, so no three-dimensional work arises.

Angles of elevation inside a solid

Put a bearing problem and a 3-D problem together and you have the standard Paper 2 question. The pieces are still the ones from 9.2 to 9.4; what changes is that the triangle you want is vertical and sits on top of a horizontal triangle you have to solve first.

Walkthrough 19 — the space diagonal of a cuboid Intermediate

The figure shows a cuboid ABCDEFGH with AB = 12 cm, BC = 9 cm and vertical edges of length 8 cm. Find (i) the length of AC, (ii) the angle of elevation of G from A and (iii) the length of AG.

In triangle ABC, ∠ABC = 90°, so AC² = 12² + 9² = 144 + 81 = 225.

ABCD is a rectangle, so its corners are right angles and Pythagoras applies in the horizontal base. This is stage 1 — and AC is asked for here, though in many versions of this question it is not.
AC = $\sqrt{225}$ = 15 cm (exact)

225 is a perfect square, so this length is exact — no rounding and no "(to 3 s.f.)". 9, 12, 15 is a Pythagorean triple, which is why the numbers came out so cleanly.
The angle of elevation of G from A is ∠GAC, and ∠ACG = 90°.

"Elevation from A" is measured from the horizontal, which here is AC, so the angle wanted sits at A in triangle ACG. The right angle is at C because CG is vertical and AC lies in the base.
tan ∠GAC = $\frac{C G}{A C}$ = $\frac{8}{15}$ so ∠GAC = 28.1° (to 1 d.p.)

Stage 2, in the redrawn vertical triangle: the rise CG is opposite and the base diagonal AC is adjacent, so tangent. The exact 15 from part (i) goes straight in, so no rounding error can enter.
AG² = 15² + 8² = 289, so AG = 17 cm (exact)

Pythagoras in the same vertical triangle. 8, 15, 17 is another triple, so the space diagonal is exact too — and a cuboid's space diagonal is always the longest straight line inside it, longer than any edge and longer than any face diagonal.

Walkthrough 20 — bearings on the ground, a mast above it Advanced

Three points A, B and C lie on level ground. B is due north of A, the bearing of C from A is 034°, the bearing of C from B is 106° and AB = 180 m. A vertical mast CT of height 46 m stands at C. Find (i) the distance BC and (ii) the angle of elevation of T from B.

Fig. 9.21

∠BAC = 34° and ∠ABC = 180° − 106° = 74°

B is due north of A, so AB is the north line at both points: at A the bearing is already the interior angle, and at B the direction to A is due south, leaving 74° of the straight line.
∠ACB = 180° − 34° − 74° = 72°

Angle sum, which also completes the side–angle pair the sine rule needs: the 72° at C sits opposite the only side given.
$\frac{B C}{\sin 34°} = \frac{180}{\sin 72°}$ so BC = 105.83… = 106 m (to 3 s.f.)

Everything so far has been in the horizontal plane, where the sine rule works exactly as it did in 9.2. BC faces the 34° and the 180 m faces the 72°.
The angle of elevation of T from B is ∠TBC, and ∠TCB = 90°.

The mast is vertical, so it is a normal to the level ground and makes a right angle with CB, which lies on that ground. Triangle TCB is therefore a flat right-angled triangle standing on BC.
tan ∠TBC = $\frac{46}{105.83…}$ so ∠TBC = 23.5° (to 1 d.p.)

The unrounded 105.83… is used, not the 106 m written in part (i) — that is the paper's "shown to a higher degree of accuracy" rule in action, and using 106 here would put the last digit of the angle at risk.

Past-paper questions

G3 N2016 · Paper 2 · Q8 12 marks

The diagram shows a solid triangular prism with three rectangular faces. AB = 7 cm, AC = 3 cm, BC = 6 cm and BD = 10 cm.

Show that angle BAC = 58.4°, correct to 1 decimal place. [3]
Calculate the surface area of the prism. [3]
Calculate the vertical distance of C above AB. [2]
Calculate the angle of elevation E from A. [4]

Worked solution

(a) Triangle ABC has all three sides known, so the cosine rule gives the angle at A:

$\cos ∠ B A C = \frac{7^{2} + 3^{2} - 6^{2}}{2 \times 7 \times 3} = \frac{22}{42} = 0.52380…$

angle BAC = 58.411…° = 58.4° (shown).

(b) The prism has two triangular ends and three rectangles, each 10 cm long:

area of one triangle $= \frac{1}{2} \times 7 \times 3 \times \sin 58.411…° = 8.9442…$ cm²

surface area = 2 × 8.9442… + 10 × (7 + 3 + 6) = 177.888… = 178 cm² (to 3 s.f.).

(c) The prism rests on the rectangular face through A and B, so the vertical distance of C above AB is the perpendicular height of triangle ABC measured from the side AB. Taking the 58.4° at A with AC = 3 as the hypotenuse,

height $= 3 \sin 58.411…° = 2.5555…$ = 2.56 cm (to 3 s.f.).

(d) Call the fourth corner of the base F, so the prism rests on the rectangle ABDF with AF = BD = 10 cm and the far triangle FDE congruent to ABC. Then E stands 2.5555… cm above the point X on FD, where

$F X = 3 \cos 58.411…° = 1.5714…$ cm

AF is perpendicular to FD, so in the flat base

${A X}^{2} = 10^{2} + {(1.5714…)}^{2} = 102.469…, A X = 10.122…$ cm

Triangle AXE is right-angled at X, so the angle of elevation is

tan (angle) $= \frac{2.5555…}{10.122…}$ , giving 14.168…° = 14.2° (to 1 d.p.).

Why this works. The prism lies on a rectangular face, so the two triangles stand up vertically and C and E are the high points of the solid — which is why (c) asks for the height of triangle ABC and (d) reuses that same 2.5555… cm. The rest of (d) is horizontal: the distance from A to the foot of E is Pythagoras on the flat base.

The greatest angle of elevation

One question type deserves its own paragraph, because it looks like it needs calculus and does not.

A mast of fixed height, seen from a moving point. With the mast at M and the observer at X on the road, tan (elevation) = height ÷ MX. The height is fixed, so the angle is greatest when MX is smallest.

So the greatest elevation is seen from the foot of the perpendicular. The shortest distance from a point to a line is the perpendicular, so the steepest view of the mast is from the point where it meets the road.

Fig. 9.22

MN is perpendicular to the road, so it is shorter than MX₁ or MX₂ or any other line from M to the road. The mast is therefore seen at its steepest from N.

The working is two lines, and the first one is a sentence. Say that the greatest angle occurs at the foot of the perpendicular; then find MN and take the inverse tangent of height ÷ MN.

More detail

The perpendicular distance itself comes from ordinary trigonometry in the horizontal plane — often MN = MP sin θ for some given point P on the road and angle θ at P.

The opening sentence carries a mark of its own: it is the statement that explains why the foot of the perpendicular is the right point, and the arithmetic that follows is only two keystrokes.

Check yourself

A cuboid has a base 7 cm by 24 cm and vertical edges of 20 cm. Find the length of a diagonal of the base, and then the length of the space diagonal.
Answer
Base diagonal: $\sqrt{7^{2} + 24^{2}}$ = $\sqrt{625}$ = 25 cm. Space diagonal: $\sqrt{25^{2} + 20^{2}}$ = $\sqrt{1025}$ = 32.015… = 32.0 cm (to 3 s.f.).
In the cuboid of the previous question, find the angle the space diagonal makes with the base.
Answer
The vertical edge is a normal to the base, so the triangle formed by the base diagonal, the vertical edge and the space diagonal is right-angled. tan θ = 20 ÷ 25 = 0.8, so θ = 38.7° (to 1 d.p.).
A student says "∠ABC looks like about 70° in the diagram, so 70° is what should go into the working." What is wrong with that in a 3-D question?
Answer
A three-dimensional solid drawn on flat paper distorts every angle that is not in the plane of the page. An angle that is really 90° can be drawn as 60° or 120°. Angles must be deduced from the solid — from a rectangular face, or from a normal to a plane — and never measured or estimated off the drawing.
A vertical mast MN stands at M on level ground, 40 m from a straight road, where N is the foot of the perpendicular from M to the road. The mast is 22 m high. Find the greatest angle of elevation of the top of the mast from a point on the road.
Answer
The elevation is greatest where the distance to M is least, which is at the foot of the perpendicular, 40 m away. So tan θ = 22 ÷ 40 = 0.55, giving θ = 28.8° (to 1 d.p.).
A pyramid has a square base WXYZ of side 10 cm and its vertex V is vertically above Y, with VY = 12 cm. Find ∠VWY.
Answer
WY is a diagonal of the square, so WY² = 10² + 10² = 200 and WY = $\sqrt{200}$ = 14.142… cm. VY is vertical and YW lies in the base, so ∠VYW = 90° and tan ∠VWY = 12 ÷ 14.142… = 0.84852…, giving 40.3° (to 1 d.p.).
Why does a 3-D question so often make you calculate a length that it never asks about?
Answer
Because the triangle containing the answer is usually vertical, and one of its sides is a diagonal of the horizontal base — a length that appears in no face of the solid and is never given. Finding it is stage 1 of the standard two-stage method; without it, stage 2 has nothing to stand on.

Which rule for which question

This is the table to know. Read down the first column until you find what the question has given you; the rest of the row is the decision.

What you are given	What you want	Use	Where the marks go
an angle and its supplement, or a sine that is not an angle	a value, or an angle	sin `A` = sin (180° − `A`), cos `A` = −cos (180° − `A`)	giving both angles when the sine is known; a negative cosine means obtuse
two sides and the included angle	the area	½`ab` sin `C`	checking the angle really is between the two sides; units²
two angles and any side	a side	Sine Rule (side over sine)	the third angle from the angle sum first, so a pair exists
two sides and a non-included angle	an angle	Sine Rule (sine over side)	writing down the supplement and the test that keeps or kills it
two sides and the included angle	the third side	Cosine Rule, `a`² = `b`² + `c`² − 2`bc` cos `A`	the square root at the end; the sign of cos `A`
all three sides	any angle	Cosine Rule rearranged	subtracting the square of the side opposite the wanted angle
a right angle is present	anything	SOH-CAH-TOA or Pythagoras	spotting it — it is always quicker than either rule
"angle of elevation / depression"	a height or a distance	a sketch, then the tangent ratio	measuring from the horizontal; moving a depression down with "alt. ∠s"
bearings	a distance or another bearing	north lines at every point, then sine or cosine rule	turning bearings into interior angles; three digits in the answer
a cuboid, a pyramid, a mast	a length or an angle	Pythagoras in the base, then a ratio in the vertical triangle	naming the right angle and why; redrawing the triangle flat
a quadrilateral, or any four-sided region	a length or an area	a diagonal, splitting it into two triangles	drawing the diagonal; adding the two areas
a moving observer and a fixed object	the greatest angle of elevation	the perpendicular distance	the sentence explaining why the perpendicular is the shortest

The formula cards

1. Given to you on the paper. Copy them from the front of the booklet; they cost nothing to look up. What the printed page cannot do is set up the lettering — in your triangle, decide which side is a first.

Area of △ A B C = \frac{1}{2} a b \sin C

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

a^{2} = b^{2} + c^{2} - 2 b c \cos A

2. Not given — the supplementary-angle pair. Same sine, opposite cosine. Everything about the ambiguous case comes from the first of these, and every "why is that negative?" from the second.

\sin A = \sin (180° - A) \cos A = - \cos (180° - A)

3. Not given — the cosine rule rearranged. The square that is subtracted is the one on the side opposite the angle you are finding. A negative result is not an error — it is an obtuse angle.

\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}

4. Not given — the bearings conventions. From north, clockwise, three digits. North 000°, east 090°, south 180°, west 270°. "The bearing of X from Y" means you are standing at Y.

5. Not given — the back bearing. The bearing of A from B differs from the bearing of B from A by 180°: add it if the result stays under 360°, subtract it otherwise.

6. Not given — the 3-D tests. A triangle in a solid is right-angled if its vertex sits at the corner of a rectangle containing all three points, or if one of its sides is a normal to a plane containing the other.

7. A vertical edge or mast is such a normal. It stands on the horizontal base, so it makes a right angle with every line drawn on that base. Read the angle off the solid rather than off the drawing.

The habits that earn the marks

1. Draw the triangle — even when a diagram is printed, redraw the one triangle you are working in, on its own, flat.
2. Label it with the question's own letters, and mark every given length and angle on it before writing any working.
3. Look for a right angle first. If there is one, SOH-CAH-TOA beats both rules.
4. Write the rule down in full before substituting into it; that line is worth a mark on its own.
5. Pair every side with the angle opposite it, and check the pairing before you press a key.
6. After a sine rule for an angle, write down the supplement and the test that keeps or kills it.
7. After a cosine rule, remember the square root.
8. Carry unrounded values from one part of a question to the next; round once, at the end of each part.
9. Lengths and areas to 3 s.f., angles in degrees to 1 d.p., bearings in three digits, areas in units².
10. Sanity-check the answer against the picture: the longest side faces the largest angle, the hypotenuse beats both other sides, and an obtuse angle makes the side opposite it longer than Pythagoras alone would.

Syllabus at a glance

The four G3 Mathematics (K310) outcomes this chapter covers, quoted from the Secondary Three/Four section of the syllabus.

G3 Mathematics (K310, first examination 2027) — these notes are written to it.

LO	What you must be able to do	Covered in
4.4	extending sine and cosine to obtuse angles	9.1 Obtuse angles
4.5	use of the formula $\frac{1}{2} a b \sin C$ for the area of a triangle	9.1 Area of a triangle
4.6	use of sine rule and cosine rule for any triangle	9.2 The Sine Rule, 9.2 The ambiguous case, 9.3 The Cosine Rule, 9.3 Choosing between them
4.7	problems in two and three dimensions including those involving angles of elevation and depression and bearings	9.4 Elevation & depression, 9.4 Bearings, 9.4 Bearings problems, 9.5 Three dimensions, 9.5 Greatest angle of elevation

Wording is quoted verbatim from the Secondary Three/Four section of the G3 Mathematics syllabus, and matches the G4 row of the K310 content table. One thing this chapter leans on is not listed in the Secondary Three/Four block, because it is met earlier: outcome 4.3, "use of trigonometric ratios (sine, cosine and tangent) of acute angles to calculate unknown sides and angles in right-angled triangles", together with 4.1 and 4.2 on Pythagoras' Theorem. Those are Secondary Two outcomes in G3, and they are recapped in 9.1. The angle between a line and a plane and the angle between two planes are in neither list and are not taught here: every three-dimensional problem in 9.5 is reduced to a right-angled triangle instead. The G2 syllabus wording for outcomes 4.4 to 4.7 is identical to the G3 wording, item for item and under the same numbers, so nothing in this chapter is marked as outside the G2 course. The only difference between the two bands is one of timing: G2 meets right-angled trigonometry (4.3) in Secondary Three/Four rather than in Secondary Two, so for a G2 reader the recap in 9.1 is current work rather than revision.

📐 Trigonometry — Revision Notes

Five questions, and the tool each one wants

What you are standing on ⓘ Recap

Past-paper questions

The lettering every formula in this chapter assumes

Sine and cosine of an obtuse angle

Past-paper questions

Area of a triangle — ½`ab` sin `C`

Check yourself

Past-paper questions

The Sine Rule

When the sine rule is the right tool

Past-paper questions

The ambiguous case — when both answers are real

Check yourself

The Cosine Rule

Finding a side: two sides and the angle between them

Finding an angle: three sides

Choosing between the two rules

Check yourself

Past-paper questions

Angles of elevation and depression

Past-paper questions

Bearings

Past-paper questions

Bearings problems — the triangle inside the compass

Check yourself

Past-paper questions

Reducing three dimensions to two

Past-paper questions

Angles of elevation inside a solid

Past-paper questions

The greatest angle of elevation

Check yourself

Which rule for which question

The formula cards

The habits that earn the marks

Syllabus at a glance

Ask CherAI

📐 Trigonometry — Revision Notes

Five questions, and the tool each one wants

What you are standing on ⓘ Recap

Past-paper questions

The lettering every formula in this chapter assumes

Sine and cosine of an obtuse angle

Past-paper questions

Area of a triangle — ½ab sin C

Check yourself

Past-paper questions

The Sine Rule

When the sine rule is the right tool

Past-paper questions

The ambiguous case — when both answers are real

Check yourself

The Cosine Rule

Finding a side: two sides and the angle between them

Finding an angle: three sides

Choosing between the two rules

Check yourself

Past-paper questions

Angles of elevation and depression

Past-paper questions

Bearings

Past-paper questions

Bearings problems — the triangle inside the compass

Check yourself

Past-paper questions

Reducing three dimensions to two

Past-paper questions

Angles of elevation inside a solid

Past-paper questions

The greatest angle of elevation

Check yourself

Which rule for which question

The formula cards

The habits that earn the marks

Syllabus at a glance

Area of a triangle — ½`ab` sin `C`