Sine Rule, Cosine Rule, and Area of a Triangle

Solving problems in non-right-angled triangles

SPEC 5.1 • TRIGONOMETRY

Learning Objectives

By the end of this lesson you will be able to:

Use the sine rule to find unknown sides and angles

Recognise and handle the ambiguous case of the sine rule

Use the cosine rule to find unknown sides and angles

Calculate the area of a triangle using ½ab sin C

Choose the appropriate rule for different problems

Non-Right-Angled Triangles

SOHCAHTOA only works in right-angled triangles. For other triangles, we need new tools:

Sine Rule

Use when you have:

A pair of opposite side and angle

Plus one more side or angle

Cosine Rule

Use when you have:

All three sides (SSS)

Two sides and included angle (SAS)

Remember

These rules work for ALL triangles, including right-angled ones!

Triangle Labelling Convention

Convention

Vertices are labelled with capital letters: A, B, C

Sides are labelled with lowercase letters: a, b, c

Side a is opposite angle A (and so on)

The Sine Rule

Sine Rule

\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C}

or equivalently:

\frac{s i n A}{a} = \frac{s i n B}{b} = \frac{s i n C}{c}

When to Use

Use the sine rule when you have a matching pair (a side and its opposite angle) plus one other piece of information.

Finding a Side Using Sine Rule

Worked Example01

In triangle ABC, angle A = 40°, angle B = 75°, and side b = 12 cm. Find the length of side a.

Step 1 Identify the matching pair: b = 12 and B = 75°

Step 2 Set up the sine rule:

\frac{a}{s i n A} = \frac{b}{s i n B}

\frac{a}{s i n 40°} = \frac{12}{s i n 75°}

Step 3 Solve for a:

a = \frac{12 \times s i n 40°}{s i n 75°} = \frac{12 \times 0.6428}{0.9659} = 7.98 cm

Finding an Angle Using Sine Rule

Worked Example02

In triangle ABC, a = 8 cm, b = 11 cm, and angle A = 35°. Find angle B.

Step 1 Set up the sine rule:

\frac{s i n B}{b} = \frac{s i n A}{a}

\frac{s i n B}{11} = \frac{s i n 35°}{8}

Step 2 Solve for sin B:

sin B = \frac{11 \times s i n 35°}{8} = \frac{11 \times 0.5736}{8} = 0.7887

Step 3 Find B:

B = sin^{- 1} (0.7887) = 52.1°

The Ambiguous Case

Warning

When using the sine rule to find an angle, there may be two possible answers!

This happens because sin θ = sin(180° − θ).

If sin B = 0.7887, then:

B = 52.1° (acute angle)

OR B = 180° − 52.1° = 127.9° (obtuse angle)

How to Check

Both answers are valid only if the triangle is still possible (all angles sum to 180°).

If the obtuse angle makes the sum exceed 180°, reject it.

The Ambiguous Case Example

Worked Example03

In triangle ABC, a = 7 cm, c = 10 cm, and angle A = 30°. Find the possible values of angle C.

Step 1 Use sine rule:

\frac{s i n C}{c} = \frac{s i n A}{a}

sin C = \frac{10 \times s i n 30°}{7} = \frac{10 \times 0.5}{7} = 0.7143

Step 2 Find both possible values:

C = sin⁻¹(0.7143) = 45.6° or C = 180° − 45.6° = 134.4°

Step 3 Check both are valid:

If C = 45.6°: A + C = 30° + 45.6° = 75.6° < 180° ✓

If C = 134.4°: A + C = 30° + 134.4° = 164.4° < 180° ✓

Both answers are valid: C = 45.6° or C = 134.4°

The Cosine Rule — Finding Sides

Cosine Rule (for sides)

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

Similarly:

b^{2} = a^{2} + c^{2} - 2 a c cos B

c^{2} = a^{2} + b^{2} - 2 ab cos C

When to Use

Use when you have two sides and the included angle (SAS).

Finding a Side Using Cosine Rule

Worked Example04

In triangle ABC, b = 8 cm, c = 11 cm, and angle A = 50°. Find the length of side a.

Step 1 Identify: We have b, c, and the included angle A → use cosine rule

Step 2 Apply the formula:

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

a^{2} = 8^{2} + 1 1^{2} - 2 (8) (11) cos 50°

a^{2} = 64 + 121 - 176 \times 0.6428

a^{2} = 185 - 113.13 = 71.87

Step 3 Take square root:

a = 71.87 = 8.48 cm

The Cosine Rule — Finding Angles

Cosine Rule (rearranged for angles)

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

Similarly:

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

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

When to Use

Use when you have all three sides (SSS).

Finding an Angle Using Cosine Rule

Worked Example05

In triangle ABC, a = 7 cm, b = 9 cm, and c = 12 cm. Find angle C.

Step 1 We have all three sides → use cosine rule for angles

Step 2 Apply the formula:

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

cos C = \frac{7 ^{2} + 9 ^{2} - 1 2 ^{2}}{2 ( 7 ) ( 9 )}

cos C = \frac{49 + 81 - 144}{126} = \frac{- 14}{126} = - 0.1111

Step 3 Find C:

C = cos^{- 1} (- 0.1111) = 96.4°

When to Use Which Rule

Given InformationRule to Use

Two angles and one side (AAS or ASA)Sine Rule

Two sides and non-included angle (SSA)Sine Rule (watch for ambiguous case!)

Two sides and included angle (SAS)Cosine Rule

Three sides (SSS)Cosine Rule

Quick Test

Do you have a matching pair (side with its opposite angle)? → Sine Rule

Otherwise → Cosine Rule

Area of a Triangle

Area Formula

Area = \frac{1}{2} ab sin C

Equivalently:

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

Remember

You need two sides and the included angle (the angle between them).

This generalises the base × height formula: ½ × base × height = ½ × a × b sin C

Finding Area

Worked Example06

In triangle ABC, b = 14 cm, c = 9 cm, and angle A = 62°. Find the area of the triangle.

Step 1 Identify: We have b, c, and included angle A

Step 2 Apply the formula:

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

Area = \frac{1}{2} \times 14 \times 9 \times sin 62°

Area = \frac{1}{2} \times 14 \times 9 \times 0.8829

Area = 55.6 cm^{2}

Finding a Side Given Area

Worked Example07

Triangle ABC has area 40 cm², side a = 10 cm, and angle B = 53°. Find side c.

Step 1 Set up the area formula:

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

40 = \frac{1}{2} \times 10 \times c \times sin 53°

Step 2 Solve for c:

40 = 5 c \times 0.7986

40 = 3.993 c

c = \frac{40}{3.993} = 10.0 cm

Multi-Step Problems

Some problems require using multiple rules in sequence:

Common Strategies

1. Use the cosine rule to find a third side, then sine rule for angles

2. Use the sine rule to find a second angle, then angles sum to 180° for third

3. Find all sides/angles first, then calculate the area

Exam Tip

Always check: do you have enough information? Look for matching pairs or SAS/SSS configurations.

Multi-Step Problem

Worked Example08

In triangle ABC, a = 5 cm, b = 7 cm, and c = 9 cm. Find the area of the triangle.

Step 1 We have SSS, so first find an angle using cosine rule:

cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab} = \frac{25 + 49 - 81}{2 ( 5 ) ( 7 )} = \frac{- 7}{70} = - 0.1

C = cos^{- 1} (- 0.1) = 95.74°

Step 2 Now use the area formula:

Area = \frac{1}{2} ab sin C

Area = \frac{1}{2} \times 5 \times 7 \times sin 95.74°

Area = \frac{1}{2} \times 5 \times 7 \times 0.995 = 17.4 cm^{2}

Practice

1 In triangle PQR, angle P = 48°, angle Q = 67°, and side p = 15 cm.

(a) Find side q.

(b) Find the area of the triangle.

Part (a) Use sine rule:

\frac{q}{s i n Q} = \frac{p}{s i n P}

q = \frac{15 \times s i n 67°}{s i n 48°} = \frac{15 \times 0.9205}{0.7431} = 18.6 cm

Part (b) First find angle R = 180° − 48° − 67° = 65°

Area = \frac{1}{2} pq sin R = \frac{1}{2} \times 15 \times 18.6 \times sin 65°

Area = \frac{1}{2} \times 15 \times 18.6 \times 0.9063 = 126.5 cm^{2}

Practice

2 In triangle XYZ, x = 6 cm, y = 8 cm, and angle Z = 120°.

(a) Find side z.

(b) Find angle X.

Part (a) Use cosine rule (SAS):

z^{2} = x^{2} + y^{2} - 2 x y cos Z

z^{2} = 36 + 64 - 2 (6) (8) cos 120°

z^{2} = 100 - 96 (- 0.5) = 100 + 48 = 148

z = 148 = 12.2 cm

Part (b) Use sine rule:

\frac{s i n X}{x} = \frac{s i n Z}{z}

sin X = \frac{6 \times s i n 120°}{12.2} = \frac{6 \times 0.866}{12.2} = 0.426

X = sin^{- 1} (0.426) = 25.2°