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CIE iGCSE Maths E6.2 Right-angled triangles Exam Style Practice Questions Paper 4 - New Syllabus

Question

(a) 

ACD is a right-angled triangle.
B is on AC and BC = 54.5 m.
AD = 83.2 m and angle ABD = 38°.

Calculate angle ACD.

(b) 

EFG is a right-angled triangle.
A circle can be drawn that passes through the three vertices of the triangle.

On the diagram, mark the position of the centre of the circle with a cross.
Explain how you decide.

(c) 

In triangle LMN, the ratio angle L : angle M : angle N = 4 : 5 : 6.
In triangle PQR, PQ = 6 cm, PR = 4 cm and QR = 5 cm.

Calculate the difference between the largest angle in triangle PQR and the largest angle in triangle LMN.

▶️ Answer/Explanation
Solution

(a) 27.3° or 27.32° to 27.33°

First find AB using tan(38°) = AD/AB, giving AB ≈ 106.5m. Then AC = AB + BC ≈ 161m.

In right triangle ACD, tan(∠ACD) = AD/AC ≈ 83.2/161, giving ∠ACD ≈ 27.3°.

(b)

Centre marked at midpoint of hypotenuse FG.

For a right-angled triangle, the hypotenuse is the diameter of the circumscribed circle, so its midpoint is the center.

(c) 10.8° or 10.81° to 10.82°

For LMN: angles are 48°, 60°, 72° (sum of ratios is 15, 180°÷15=12° per part). Largest is 72°.

For PQR: using cosine rule on largest angle opposite longest side (6cm), angle ≈ 82.8°.

Difference: 82.8° – 72° = 10.8°.

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