The area of a triangle is ½ab sin C, but only when C is the angle between sides a and b. When the given angle sits elsewhere, you find one more piece of information first.
This lesson is part of solution of triangles. It uses the sine rule from choosing sine rule or cosine rule.
Why is the included angle needed?
The formula ½ab sin C is really ½ × base × height. Take side a as the base, and the height from the opposite corner is b sin C, because b and C together make a right-angled triangle with that height.
That only works when C lies between a and b. With any other angle, b sin(angle) is not the height above the base.
Worked example 1: a missing side first
In triangle PQR, PQ = 10 cm, angle P = 50° and angle Q = 60°. Find the area.
Only one side is known, so the area formula cannot be used yet. Find a second side first.
- Angle R = 180° − 50° − 60° = 70°.
- Sine rule: PR ÷ sin 60° = 10 ÷ sin 70°, so PR = 10 × 0.8660 ÷ 0.9397 ≈ 9.216 cm.
- Now PQ and PR enclose angle P = 50°, so the area = ½ × 10 × 9.216 × sin 50° = 46.08 × 0.7660 ≈ 35.3 cm².
Worked example 2: the same area from a height
Check the answer with a height. Drop a perpendicular from R to PQ. In the right-angled triangle formed at P, the height h = PR sin 50° = 9.216 × 0.7660 ≈ 7.060 cm.
Area = ½ × 10 × 7.060 = 35.3 cm², the same value. Two routes agreeing is a strong check.
The mistake that costs marks
The common slip is to pair the given angle with whichever two sides are nearby. Using example 1, a student might compute ½ × 10 × 9.216 × sin 60°, which uses the angle at Q although Q is not between PQ and PR.
| Step | Wrong | Right |
|---|---|---|
| Choose the two sides | PQ and PR | PQ and PR |
| Choose the angle | 60° (angle Q) | 50° (angle P, between them) |
| Result | 39.9 cm² | 35.3 cm² |
The test takes five seconds: put a finger on the two sides you use, and confirm the angle is the corner where they meet.
When a height is asked for directly
Sometimes the question asks for the distance from a corner to a side. That distance is the height, and a sine ratio gives it. In example 1 the distance from R to PQ is 7.06 cm.
If a height is already given, use it directly: area = ½ × base × height, where the height is perpendicular to the base. Check the units, then round at the end.
Check yourself
In triangle ABC, AB = 12 cm, angle A = 65° and angle B = 55°. Find the area.
Answer
Angle C = 180° − 65° − 55° = 60°.
Sine rule: AC ÷ sin 55° = 12 ÷ sin 60°, so AC = 12 × 0.8192 ÷ 0.8660 ≈ 11.351 cm.
AB and AC enclose angle A = 65°, so area = ½ × 12 × 11.351 × sin 65° = 68.10 × 0.9063 ≈ 61.7 cm².
What to study next
Area questions often sit inside larger diagrams. Continue with solving three-dimensional triangle applications, where you pick a flat triangle first. The word-problem structure worksheet helps you organise a long diagram question.
If you want a teacher to work through these with you, see online one-to-one Additional Mathematics tuition.