Lesson 7: Trigonometry

Trigonometry — the mathematics of triangles — lets you measure what you cannot reach: the height of Blackpool Tower, the distance to a ship off the Cornish coast, the angle of a wheelchair ramp in a Leeds shopping centre. This lesson covers SOHCAHTOA for right-angled triangles, the exact values, and the Higher-tier sine and cosine rules for triangles without a right angle. Trigonometry appears on both GCSE papers, so build both calculator fluency and non-calculator technique.

SOHCAHTOA — the three ratios

In a right-angled triangle, label the sides relative to the angle θ: the hypotenuse (opposite the right angle, always longest), the opposite (across from θ), and the adjacent (next to θ, not the hypotenuse). The ratios:

sin θ = opposite / hypotenuse
cos θ = adjacent / hypotenuse
tan θ = opposite / adjacent

The mnemonic SOH-CAH-TOA is pronounced 'sock-a-toe-a'. Finding a side: if you know the angle and one side, pick the ratio containing both knowns and the unknown. Worked example: in a right-angled triangle, the hypotenuse is 15 cm and the angle is 40°. Find the opposite side. sin 40° = opp/15, so opp = 15 × sin 40° = 15 × 0.6428 = 9.64 cm.

Finding an angle: use the inverse functions (sin⁻¹, cos⁻¹, tan⁻¹ — the SHIFT key). Worked example: opposite 7 cm, adjacent 11 cm. tan θ = 7/11 = 0.6364, so θ = tan⁻¹(0.6364) = 32.5°. Always check the answer is plausible: angles in a triangle are less than 180°, and in a right-angled triangle the two acute angles sum to 90°.

Choosing the ratio: write down what you have and what you want ('I have H and A, I want the angle → cos'). The labelling step — especially spotting the hypotenuse — is where most errors happen, so do it on the diagram every time. Non-calculator questions use the standard angles (30°, 45°, 60°) with exact values, below.

Angles of elevation and depression

The angle of elevation is measured up from the horizontal; the angle of depression is measured down from the horizontal — and by alternate angles, the angle of depression from A to B equals the angle of elevation from B to A. This trick converts awkward depression problems into standard ones.

Worked example: Blackpool Tower is 158 m tall. How far from the base are you standing if the angle of elevation of the top is 12°? tan 12° = 158/d, so d = 158 ÷ tan 12° = 158 ÷ 0.2126 = 743 m. (Sanity check: a shallow angle means a long distance — 743 m for a 12° angle is plausible.)

Worked example: a lighthouse keeper 40 m above sea level sees a boat at an angle of depression of 8°. The horizontal distance to the boat is 40 ÷ tan 8° = 40 ÷ 0.1405 = 285 m. Draw the horizontal line at the keeper, mark the 8° below it, and transfer it to the boat's angle of elevation — the Z pattern from Lesson 5.

Worked example (ramps and gradients): a wheelchair ramp rises 0.6 m over a horizontal distance of 7.2 m. Its angle is tan⁻¹(0.6/7.2) = tan⁻¹(0.0833) = 4.8°. UK building regulations allow a maximum gradient of 1 in 12 (≈ 4.8°) for short ramps — so this ramp is right at the limit. Trigonometry is how accessibility is engineered.

Exact values — the non-calculator triangles (Higher)

Two special triangles generate the exact values you must know: the isosceles right-angled triangle (legs 1, 1; hypotenuse √2) gives angles of 45°; half an equilateral triangle (sides 1, 2; height √3) gives 30° and 60°. The table:

• sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
• sin 45° = 1/√2, cos 45° = 1/√2, tan 45° = 1
• sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3

Worked example: find the exact value of sin 45° × tan 60° = (1/√2) × √3 = √3/√2 = √6/2 (rationalising the denominator by multiplying top and bottom by √2). Answers in surd form are exact — never replace √3 with 1.73 in an exact-value question. (Deriving these from the two triangles is itself a favourite one-mark question.)

The sine rule and cosine rule (Higher)

For triangles without a right angle, two rules take over. The sine rule: a/sin A = b/sin B = c/sin C — use it when you know a side-angle opposite pair. The cosine rule: a² = b² + c² − 2bc cos A — use it to find a side from two sides and the included angle, or an angle from three sides (rearranged: cos A = (b² + c² − a²)/2bc).

Worked example (sine rule): in triangle PQR, angle P = 52°, angle Q = 71°, and side p (opposite P) = 10 cm. Find q. First angle R = 180° − 52° − 71° = 57°. Then q/sin 71° = 10/sin 52°, so q = 10 × sin 71° ÷ sin 52° = 10 × 0.9455 ÷ 0.7880 = 12.0 cm.

Worked example (cosine rule): two walkers leave a Lake District car park on bearings differing by 50°; one walks 6 km, the other 8 km. How far apart are they? The angle between their paths is 50°, so d² = 6² + 8² − 2(6)(8) cos 50° = 36 + 64 − 96 × 0.6428 = 100 − 61.7 = 38.3, giving d = 6.19 km.

Area of a triangle: area = ½ab sin C — use it when you know two sides and the included angle. For sides 7 and 9 with included angle 64°: area = ½ × 7 × 9 × sin 64° = 31.5 × 0.8988 = 28.3 cm². (When a base and perpendicular height are known, the familiar ½ × base × height still applies — it is the special case where sin C = sin 90° = 1.)

A caution on the sine rule for angles: finding an angle from sin θ = 0.5 gives θ = 30° or 150° (the ambiguous case, since sin(180° − θ) = sin θ). GCSE rarely tests the ambiguity directly, but if your angle makes the triangle impossible (angles summing past 180°), try the supplement.

Problem solving with trigonometry

Real problems combine trigonometry with bearings (Lesson 5) and Pythagoras. Strategy: draw the diagram, mark the north lines and known sides/angles, split the problem into right-angled triangles if possible, and choose rules by what you know: right angle → SOHCAHTOA; opposite pair → sine rule; included angle or three sides → cosine rule.

Worked example: from a point A, a church spire's top has angle of elevation 23°. Walking 80 m closer to point B, the angle is 35°. Let the height be h and AB = 80. From B: h = x tan 35° where x is the distance from B. From A: h = (x + 80) tan 23°. So x tan 35° = (x + 80) tan 23° → x(0.7002) = (x + 80)(0.4245) → 0.7002x − 0.4245x = 33.96 → x = 33.96 ÷ 0.2757 = 123.2 m. Then h = 123.2 × tan 35° = 86.3 m. Two-equation problems like this are classic Higher-tier fare — the method is always 'write both expressions for h and equate'.

Problem solving — a cable across a river

Worked example: engineers must run a cable from a point on one bank of a river to a tower on the other side. The river is 60 m wide, and the tower stands 45 m along the far bank from the point directly opposite. The cable runs in a straight line. Length: Pythagoras gives √(60² + 45²) = √(3,600 + 2,025) = √5,625 = 75 m — a 3-4-5 triangle scaled by 15, which you should spot before reaching for the calculator. The angle the cable makes with the near bank: tan θ = 45/60 = 0.75 → θ = 36.9°. If cable costs £84 per metre, the crossing costs 75 × 84 = £6,300. Pythagoras and trigonometry are the same triangle read two ways — lengths via Pythagoras, angles via the trig ratios.

Worked example: a ladder 5 m long leans against a wall in a Cardiff terrace, foot 1.8 m from the wall. Height reached: √(25 − 3.24) = √21.76 ≈ 4.67 m. Angle with the ground: cos θ = 1.8/5 = 0.36 → θ = 68.9°. Safety guidance says the angle should be about 75° — this ladder is too shallow. Note which ratio appeared: the known side (1.8) is adjacent and the ladder is the hypotenuse, so cos — choosing the ratio from the labels, again.

Common mistakes

• Wrong ratio. Labelling the sides wrongly (especially the hypotenuse, or opposite/adjacent from the wrong angle) ruins everything. Label first, calculate second.
• Calculator in radian mode. GCSE works in degrees. If sin 30° gives −0.988, your calculator is in radians — switch it and redo.
• Using SOHCAHTOA on a non-right-angled triangle. No right angle means sine/cosine rules (Higher) — or split the triangle with a perpendicular.
• Premature rounding. Keep at least 4 significant figures in intermediate steps; round only the final answer (usually to 3 s.f.).
• Forgetting the inverse. Finding an angle needs sin⁻¹/cos⁻¹/tan⁻¹ — writing θ = tan(7/11) instead of tan⁻¹ is a classic slip.
• Exact-value questions answered in decimals. √3 is the answer; 1.73 is an approximation and loses the mark.

Exam Experience

Question 1. Find the exact value of sin 45° × tan 60°. Give your answer in its simplest surd form. [2 marks]

Question 2. A flagpole is vertical. From a point 20 m from the foot of the flagpole, the angle of elevation of the top of the flagpole is 31°. Work out the height of the flagpole. Give your answer to 3 significant figures. [2 marks]

Question 3. In triangle XYZ, XY = 9 cm, YZ = 7 cm and angle XYZ = 80°. Work out the length of XZ. Give your answer to 3 significant figures. [3 marks]

Mark scheme.

• Q1: M1 for the correct exact values (sin 45° = 1/√2 and tan 60° = √3). A1 for √6/2 (accept √3/√2 followed by rationalisation).
• Q2: M1 for tan 31° = h/20, so h = 20 tan 31°. A1 for 12.0 m (20 × 0.6009 = 12.02).
• Q3: M1 for the cosine rule: XZ² = 9² + 7² − 2 × 9 × 7 × cos 80°. M1 for evaluating: 130 − 126 × 0.17365 = 130 − 21.9 = 108.1. A1 for XZ = 10.4 cm (√108.1 = 10.397).