Lesson 7: Trigonometry | O Level Mathematics

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1. Trigonometry in Right-Angled Triangles

Trigonometry studies the relationships between the sides and angles of triangles. For right-angled triangles, three ratios — sine, cosine, and tangent — connect any angle to two sides.

Labelling a Right-Angled Triangle: Always label sides relative to the angle you are working with (θ).
• Hypotenuse (H) — longest side, always opposite the right angle
• Opposite (O) — side directly opposite angle θ
• Adjacent (A) — side next to angle θ (not the hypotenuse)

SOH CAH TOA — The Three Trigonometric Ratios:

sin θ = Opposite / Hypotenuse    (SOH)
cos θ = Adjacent / Hypotenuse    (CAH)
tan θ = Opposite / Adjacent      (TOA)

Finding a Side

📐 Worked Example 1 — Finding an Unknown Side

In triangle ABC, angle B = 90°, angle A = 38°, AB = 14 cm. Find BC and AC.

Label sides relative to angle A = 38°:
AB = Adjacent = 14 cm. BC = Opposite. AC = Hypotenuse.

Find BC (Opposite): Use TOA.
tan 38° = BC/14
BC = 14 × tan 38° = 14 × 0.7813 = 10.9 cm

Find AC (Hypotenuse): Use CAH.
cos 38° = 14/AC
AC = 14/cos 38° = 14/0.7880 = 17.8 cm

Finding an Angle

📐 Worked Example 2 — Finding an Unknown Angle

A right-angled triangle has sides 5 cm, 12 cm, and 13 cm. Find the smallest angle.

The smallest angle is opposite the shortest side (5 cm). Hypotenuse = 13 cm (longest).
Opposite = 5, Hypotenuse = 13 → use SOH.

sin θ = 5/13 = 0.3846
θ = sin⁻¹(0.3846) = 22.6°

Check: Other acute angle = 90° − 22.6° = 67.4°. All three: 22.6° + 67.4° + 90° = 180° ✓

Angles of Elevation and Depression

Angle of Elevation

The angle measured upward from the horizontal to a line of sight toward an object above.

Example: Looking up at the top of a tower from ground level.

Angle of Depression

The angle measured downward from the horizontal to a line of sight toward an object below.

Example: Looking down from a cliff top to a boat at sea.

📐 Worked Example 3 — Angle of Elevation

From a point 50 m from the base of a vertical tower, the angle of elevation of the top is 32°. Find the height of the tower.

Draw a right-angled triangle: horizontal distance = 50 m (Adjacent), height = h (Opposite), angle = 32°.

Use TOA:
tan 32° = h/50
h = 50 × tan 32° = 50 × 0.6249 = 31.2 m

Special Angles — Exact Values

Angle θ	sin θ	cos θ	tan θ
0°	0	1	0
30°	½	√3/2	1/√3 = √3/3
45°	1/√2 = √2/2	1/√2 = √2/2	1
60°	√3/2	½	√3
90°	1	0	undefined

Memory Trick for Exact Values:
sin 0°, 30°, 45°, 60°, 90° = √0/2, √1/2, √2/2, √3/2, √4/2 = 0, ½, √2/2, √3/2, 1
For cos, read the sin values in reverse order.

2. The Sine Rule

The sine rule applies to any triangle — not just right-angled ones. It connects each side to the sine of its opposite angle.

Sine Rule

a/sin A = b/sin B = c/sin C

or equivalently: sin A/a = sin B/b = sin C/c

Use to find a side: put the unknown side on top. Use to find an angle: put the unknown angle (as sin) on top.

When to Use the Sine Rule:
✓ Given two angles and one side (AAS or ASA)
✓ Given two sides and a non-included angle (SSA — the ambiguous case)
✗ Do NOT use when you have the included angle between two known sides → use Cosine Rule instead.

📐 Worked Example 4 — Sine Rule (Finding a Side)

In triangle ABC: angle A = 48°, angle B = 67°, side a = 15 cm. Find side b.

Write the sine rule with the unknown on top:
b/sin B = a/sin A

Substitute values:
b/sin 67° = 15/sin 48°
b = 15 × sin 67°/sin 48° = 15 × 0.9205/0.7431 = 18.6 cm

📐 Worked Example 5 — Sine Rule (Finding an Angle)

In triangle PQR: p = 10 cm, q = 14 cm, angle P = 34°. Find angle Q.

Write the sine rule with the unknown angle on top:
sin Q/q = sin P/p
sin Q/14 = sin 34°/10

sin Q = 14 × sin 34°/10 = 14 × 0.5592/10 = 0.7829
Q = sin⁻¹(0.7829) = 51.6°

Check for ambiguous case: Since q > p (14 > 10), angle Q > angle P. 51.6° > 34° ✓. The obtuse possibility: Q = 180° − 51.6° = 128.4°. Check: P + Q = 34° + 128.4° = 162.4° < 180° → this is also valid. Both solutions should be stated unless the context rules one out.

⚠ The Ambiguous Case (SSA): When using the sine rule to find an angle, there may be TWO possible answers because sin θ = sin(180° − θ). Always check whether both the acute and obtuse angles are valid. In Cambridge exams, state both if both are geometrically possible.

3. The Cosine Rule

Cosine Rule — Two Forms

a² = b² + c² − 2bc cos A (finding a side)

cos A = (b² + c² − a²) / 2bc (finding an angle)

The formula works for any vertex — just relabel: b²=a²+c²−2ac cos B, etc.

When to Use the Cosine Rule:
✓ Given two sides and the included angle (SAS) — finding the third side
✓ Given all three sides (SSS) — finding any angle
✗ Do NOT use when you have two angles and a side → use Sine Rule instead.

📐 Worked Example 6 — Cosine Rule (Finding a Side)

In triangle ABC: b = 7 cm, c = 11 cm, angle A = 55°. Find side a.

Apply the cosine rule:
a² = b² + c² − 2bc cos A
a² = 49 + 121 − 2(7)(11) cos 55°

a² = 170 − 154 × 0.5736
a² = 170 − 88.33 = 81.67
a = √81.67 = 9.04 cm

📐 Worked Example 7 — Cosine Rule (Finding an Angle)

A triangle has sides a = 8 cm, b = 11 cm, c = 6 cm. Find angle C.

Use the angle form (c is smallest side → C is smallest angle):
cos C = (a² + b² − c²) / 2ab
cos C = (64 + 121 − 36) / (2 × 8 × 11)

cos C = 149/176 = 0.8466
C = cos⁻¹(0.8466) = 32.1°

Note: When using the cosine rule to find an angle, there is NO ambiguous case — cos⁻¹ always gives a unique answer between 0° and 180°.

4. Area of a Triangle Using Sine

Area of Any Triangle

Area = ½ ab sin C

where a and b are two sides and C is the included angle between them.

This works for any triangle — not just right-angled.

📐 Worked Example 8 — Area Using Sine

Find the area of triangle PQR where PQ = 9 cm, PR = 13 cm, and angle P = 72°.

Identify the two sides (a=9, b=13) and included angle (C=72°):
Area = ½ × 9 × 13 × sin 72°

= ½ × 117 × 0.9511
= 55.6 cm²

Choosing the Right Formula — Decision Guide

Given Information	Find	Use
Right-angled triangle — any two of O, A, H	Side or angle	SOH CAH TOA
Two angles + one side (AAS/ASA)	A side	Sine Rule
Two sides + non-included angle (SSA)	An angle	Sine Rule (check ambiguous case)
Two sides + included angle (SAS)	Third side	Cosine Rule (a²=b²+c²−2bc cosA)
Three sides (SSS)	An angle	Cosine Rule (cosA=(b²+c²−a²)/2bc)
Two sides + included angle (SAS)	Area	Area = ½ab sin C
Base and perpendicular height known	Area	Area = ½ × base × height

5. Bearings

Bearing: A direction measured as an angle clockwise from North. Always written as a three-digit number (e.g. 045°, 270°, 135°). Bearings range from 000° to 360°.

Key Bearing Rules:
• Always measure clockwise from North
• Always use three digits: write 090°, not 90°
• North = 000°, East = 090°, South = 180°, West = 270°
• Back bearing: If bearing from A to B is θ, then bearing from B to A = θ + 180° (if θ < 180°) or θ − 180° (if θ > 180°)

📐 Worked Example 9 — Bearings with Trigonometry

A ship sails from port P on a bearing of 065° for 120 km to reach point Q. Find how far Q is (a) north of P (b) east of P.

The bearing 065° means 65° east of north. Draw a right-angled triangle with hypotenuse = 120 km and angle at P = 65° from North.

North component (adjacent to 65°, from North):
North = 120 × cos 65° = 120 × 0.4226 = 50.7 km

East component (opposite to 65° from North):
East = 120 × sin 65° = 120 × 0.9063 = 108.8 km

📐 Worked Example 10 — Finding a Bearing

Town B is 8 km east and 15 km north of town A. Find the bearing of B from A.

Draw a right-angled triangle: North = 15 km, East = 8 km. The bearing is the angle measured clockwise from North.

Find the angle from North:
tan θ = 8/15 (East/North)
θ = tan⁻¹(8/15) = tan⁻¹(0.5333) = 28.1°

Bearing of B from A = 028° (clockwise from North, 3 digits)

6. Trigonometry in Three Dimensions

3D trigonometry problems involve finding angles and lengths within three-dimensional solids — cuboids, pyramids, and wedges. The key technique is to identify a right-angled triangle within the 3D shape and solve it using standard trigonometry.

Strategy for 3D Trigonometry:
1. Draw a clear diagram of the 3D solid and label all known measurements.
2. Identify the right-angled triangle that contains the unknown length or angle.
3. Sometimes you need to find an intermediate length first (using another right-angled triangle).
4. Apply SOH CAH TOA, sine rule, or Pythagoras to the identified triangle.

📐 Worked Example 11 — Angle in a Cuboid

A cuboid has length 8 cm, width 6 cm, and height 5 cm. Find (a) the length of the space diagonal AG (b) the angle AG makes with the base ABCD.

Step 1 — Find diagonal of base AC:
AC = √(8² + 6²) = √(64 + 36) = √100 = 10 cm

Step 2 — Find space diagonal AG:
Triangle ACG has a right angle at C, with AC = 10 cm and CG = height = 5 cm.
AG = √(AC² + CG²) = √(100 + 25) = √125 = 5√5 ≈ 11.2 cm

Step 3 — Angle with base:
Angle GAC: opposite = CG = 5, adjacent = AC = 10.
tan(∠GAC) = 5/10 = 0.5
∠GAC = tan⁻¹(0.5) = 26.6°

📐 Worked Example 12 — Pyramid

A square-based pyramid has base side 10 cm and vertical height 12 cm. Find (a) the slant height from apex to midpoint of base edge (b) the angle the slant face makes with the base.

(a) Slant height: The right triangle has base = half of 10 cm = 5 cm (from centre of base to midpoint of edge), and vertical height = 12 cm.
l = √(5² + 12²) = √(25 + 144) = √169 = 13 cm

(b) Angle with base: In the same right triangle, opposite = 12, adjacent = 5.
tan θ = 12/5 = 2.4
θ = tan⁻¹(2.4) = 67.4°

Common 3D Mistake: Students often try to work in 3D directly. Always reduce the problem to a series of 2D right-angled triangles. Identify the triangle containing your unknown, check which sides you know, then choose the right trigonometric method.

7. Graphs of Trigonometric Functions

Understanding the shape and key features of trig graphs is required for reading off solutions graphically.

Function	Period	Range	Key Features
y = sin x	360°	−1 ≤ y ≤ 1	Starts at (0,0). Maximum 1 at 90°. Returns to 0 at 180°. Minimum −1 at 270°. Returns to 0 at 360°.
y = cos x	360°	−1 ≤ y ≤ 1	Starts at (0,1). Falls to 0 at 90°. Minimum −1 at 180°. Returns to 0 at 270°. Back to 1 at 360°.
y = tan x	180°	All real numbers	Vertical asymptotes at 90°, 270°. Passes through (0,0), (45°,1), (135°,−1). Undefined at 90° and 270°.

Reading Trig Graph Solutions:
If sin θ = k has one solution θ₁ in 0° to 90°, then in 0° to 360°:
• sin θ = k → θ = θ₁ OR θ = 180° − θ₁
• cos θ = k → θ = θ₁ OR θ = 360° − θ₁
• tan θ = k → θ = θ₁ OR θ = θ₁ + 180°

📐 Worked Example 13 — Solving Trig Equations

Solve for 0° ≤ x ≤ 360°: (a) sin x = 0.6 (b) cos x = −0.3 (c) tan x = −1

(a) sin x = 0.6:
Principal value: x = sin⁻¹(0.6) = 36.9°
Second solution: x = 180° − 36.9° = 143.1°
x = 36.9° or 143.1°

(b) cos x = −0.3:
Principal value: cos⁻¹(0.3) = 72.5°. Since cos is negative → 2nd and 3rd quadrants.
x = 180° − 72.5° = 107.5° and x = 180° + 72.5° = 252.5°
x = 107.5° or 252.5°

(c) tan x = −1:
tan⁻¹(1) = 45°. Tan is negative in 2nd and 4th quadrants.
x = 180° − 45° = 135° and x = 360° − 45° = 315°
x = 135° or 315°

📝 Exam Practice Questions

Q1 [3 marks] — A ladder 6.5 m long leans against a vertical wall. The foot of the ladder is 2.5 m from the base of the wall on horizontal ground. Find (a) the angle the ladder makes with the ground (b) how high up the wall the ladder reaches.

(a) cos θ = 2.5/6.5 → θ = cos⁻¹(0.3846) = 67.4°

(b) h = √(6.5² − 2.5²) = √(42.25 − 6.25) = √36 = 6 m
OR: h = 6.5 × sin 67.4° = 6.5 × 0.9231 = 6 m ✓

Q2 [4 marks] — In triangle ABC, AB = 12 cm, BC = 9 cm, angle ABC = 110°. Find (a) AC (b) angle BAC. Give answers to 1 decimal place.

(a) Using cosine rule: a² = b² + c² − 2bc cosA. Here finding AC (side b) with angle B = 110°:
AC² = AB² + BC² − 2(AB)(BC)cos B
AC² = 144 + 81 − 2(12)(9)cos 110°
AC² = 225 − 216(−0.3420) = 225 + 73.87 = 298.87
AC = 17.3 cm

(b) Using sine rule: sin(BAC)/BC = sin B/AC
sin(BAC) = 9 × sin 110°/17.3 = 9 × 0.9397/17.3 = 0.4886
Angle BAC = 29.3°

Exam Tip: When the given angle is obtuse (110°), cos 110° is negative — this increases the value of the third side (makes sense geometrically: the side opposite an obtuse angle is the longest). Always check your answer is the largest side when the opposite angle is obtuse.

Q3 [3 marks] — Find the area of triangle PQR where PQ = 7.4 cm, QR = 9.2 cm, and angle PQR = 64°. Find PR using the cosine rule.

Area = ½ × PQ × QR × sin(PQR) = ½ × 7.4 × 9.2 × sin 64°
= ½ × 68.08 × 0.8988 = 30.6 cm²

PR² (cosine rule): = 7.4² + 9.2² − 2(7.4)(9.2)cos 64°
= 54.76 + 84.64 − 136.16 × 0.4384 = 139.4 − 59.7 = 79.7
PR = 8.93 cm

Q4 [3 marks] — A ship sails from A to B on a bearing of 125°, a distance of 60 km. It then sails from B to C on a bearing of 215°, a distance of 45 km. Find the distance AC and the bearing of C from A.

Angle ABC: Bearing AB = 125°, bearing BC = 215°.
Back bearing of BA = 125° − 180° = 305°. Nope: back bearing = 125°+180°=305°.
Angle at B between BA and BC = 305° − 215° = 90°! Triangle ABC is right-angled at B.
AC = √(60² + 45²) = √(3600+2025) = √5625 = 75 km

Angle BAC = tan⁻¹(45/60) = 36.9°
Bearing of A to B = 125°, so bearing of C from A = 125° + 36.9° = 161.9° ≈ 162°

Q5 [4 marks] — A vertical pole PQ stands at corner P of a horizontal rectangular field PQRS where PQ = 8 m (the pole), PS = 15 m and PQ (field side) = 20 m. (Clarifying: pole height = 8 m at corner P; field is 15 m × 20 m.) Find (a) the distance from P to the far corner R (b) the angle of elevation of the top of the pole from R.

(a) Distance PR (diagonal of rectangle):
PR = √(15² + 20²) = √(225 + 400) = √625 = 25 m

(b) Angle of elevation from R to top of pole:
Right triangle: horizontal = PR = 25 m, vertical = pole height = 8 m
tan θ = 8/25 = 0.32
θ = tan⁻¹(0.32) = 17.7°

Exam Tip: 3D trig always requires finding an intermediate distance first (here the diagonal of the base). This is the most commonly missed step. Always ask: "Do I need another triangle before I can solve for the angle?"

Q6 [3 marks] — Solve for 0° ≤ x ≤ 360°: 2 cos x + 1 = 0

Working:
2 cos x = −1 → cos x = −0.5
Principal value: cos⁻¹(0.5) = 60°. Cosine is negative → 2nd and 3rd quadrants.
x = 180° − 60° = 120° and x = 180° + 60° = 240°
x = 120° or x = 240°

Exam Tip: Always rearrange first to isolate the trig function (here: cos x = −0.5), then find the principal value, then use the quadrant rules. Never forget both solutions in 0° to 360°.

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