When a triangle has two known sides and the angle between them, its area is ½ × a × b × sin C. You do not need the perpendicular height, and the rule works whether or not the triangle has a right angle.
This lesson belongs after choosing the sine or cosine rule. Check your syllabus tier on the Cambridge page, because it is Extended content.
Where does the formula come from?
The usual area formula is ½ × base × height. In a triangle with sides a and b meeting at angle C, take a as the base. The height from the far end of side b down to the base is b × sin C.
So area = ½ × a × (b sin C) = ½ab sin C. The angle C is the angle between a and b, which is why it can be read straight from the diagram.
How to use it
- Identify the two sides and check that the angle lies between them.
- Substitute into ½ab sin C.
- Give units squared (cm², m²), and round at the end.
To find an unknown angle, rearrange: sin C = 2 × area ÷ (ab). Then use the inverse sine.
Worked example 1: find the area
Two sides of a triangle are 9.4 cm and 6.8 cm, and the angle between them is 57°.
Step 1: a = 9.4, b = 6.8, C = 57°. The angle is between the two sides.
Step 2: Area = ½ × 9.4 × 6.8 × sin 57° = 31.96 × 0.8387 = 26.80.
Answer: 26.8 cm² (3 s.f.).
Worked example 2: find an angle
A triangle has area 40 cm² and two sides of length 10 cm and 12 cm. The angle between them is acute. Find it.
Step 1: 40 = ½ × 10 × 12 × sin C = 60 sin C.
Step 2: sin C = 40 ÷ 60 = 0.6667.
Step 3: C = sin⁻¹(0.6667) = 41.81°.
Answer: 41.8°. The word “acute” matters. If the angle were obtuse, it would be 180° − 41.81° = 138.2°, which has the same sine.
The mistake to watch for
A common error is to use an angle that is not between the two sides.
Mistaken working: In triangle ABC, AB = 9 cm, BC = 12 cm and angle A = 40°. The student wrote area = ½ × 9 × 12 × sin 40° = 34.7 cm².
Angle A lies between AB and AC, not between AB and BC.
The correct route finds the angle between AB and BC, which is angle B. First, sin C = 9 × sin 40° ÷ 12 = 0.4821, so C = 28.8°. Then B = 180° − 40° − 28.8° = 111.2°.
Now area = ½ × 9 × 12 × sin 111.2° = 54 × 0.9325 = 50.4 cm².
The mistaken value of 34.7 cm² is far below the correct one, so a rough check (is the area plausible for a triangle with sides 9 and 12?) would have caught the problem.
Check yourself
1. Two sides are 12 cm and 15 cm, with an angle of 30° between them. Find the area.
Show answer
Area = ½ × 12 × 15 × sin 30° = 90 × 0.5 = 45 cm².
2. A triangle has area 35 cm² and two sides of 8 cm and 14 cm. The included angle is acute. Find it.
Show answer
35 = ½ × 8 × 14 × sin C = 56 sin C. So sin C = 0.625 and C = 38.7° (3 s.f.).
3. An equilateral triangle has sides of 6 cm. Find its area.
Show answer
Each angle is 60°. Area = ½ × 6 × 6 × sin 60° = 18 × 0.8660 = 15.59, giving 15.6 cm².
Where this leads next
The obtuse partner angle in worked example 2 is the start of the next lesson: explaining an ambiguous diagram before calculating. You can also return to two-stage journey diagrams and find the area of the triangle formed. The non-calculator working trainer helps with exact values such as sin 30°, and the mixed practice set tests the whole module.
Area questions often hide a second step, such as finding a missing angle first. A teacher in online one-to-one Mathematics tuition can help you spot it before you substitute.