KCSE 2025 Maths P1 Q15 — Trigonometry: Finding a Side Using Area and Tangent

KCSE 2025 Form 4 Trigonometry

Published

The Question

“Triangle ABC is right-angled at B with AB = 8 cm. D is a point on side AC such that AD = 5 cm, and the area of the smaller triangle ABD is 10 square centimetres. Calculate the length of side BC.”

1

Spot the shared angle

Point D lies on the line AC, so the angle at vertex A is exactly the same for the small triangle ABD and the big triangle ABC. This shared angle is the bridge between the two triangles: whatever we learn about angle A from the small triangle applies directly to the large one. So the plan is to first pin down angle A using the area we are given.

2

Write the area of triangle ABD using two sides and the included angle

The area of any triangle equals one half times two of its sides times the sine of the angle sitting between them. In triangle ABD the two sides that meet at A are AB = 8 cm and AD = 5 cm, and the angle between them is angle A. Setting this equal to the given area of 10 square centimetres gives an equation with only angle A unknown.

Area=12×AB×AD×sin⁡A\text{Area} = \frac{1}{2} \times AB \times AD \times \sin A
10=12×8×5×sin⁡A=20sin⁡A10 = \frac{1}{2} \times 8 \times 5 \times \sin A = 20 \sin A
3

Solve for angle A

Rearranging the equation makes the sine of angle A the subject. Dividing both sides by 20 shows that the sine of angle A is one half. The angle whose sine is one half is 30 degrees, so angle A is 30 degrees.

sin⁡A=1020=12\sin A = \frac{10}{20} = \frac{1}{2}
A=30∘A = 30^\circ
4

Apply the tangent ratio in the big right-angled triangle

Now switch to triangle ABC, which is right-angled at B. Looking from angle A, the side AB = 8 cm is next to the angle (adjacent) and the side BC we want is across from the angle (opposite). The tangent of an angle is opposite over adjacent, so tan A equals BC over AB. Multiplying through by AB isolates BC.

tan⁡A=BCAB⇒BC=AB×tan⁡A\tan A = \frac{BC}{AB} \Rightarrow BC = AB \times \tan A
BC=8×tan⁡30∘=4.62 cmBC = 8 \times \tan 30^\circ = 4.62 \text{ cm}

Final Result

The length of side BC is 4.62 cm (to 2 decimal places).

Why this method works

The trick is that the small triangle and the large triangle share the angle at A, so information from one flows into the other. The area formula 'one half times two sides times the sine of the included angle' is powerful because it links a triangle's area to an angle, letting us extract an angle we could not measure directly. Once angle A is known, the big triangle is an ordinary right-angled triangle, so the basic tangent ratio (opposite over adjacent) finishes the job.

✓

Reverse-check: with BC = 4.62 and AB = 8, tan A = 4.62/8 = 0.577, which gives A = 30 degrees, matching the angle found from the area.