Nanyang 2025 P6 Maths Prelim — Paper 1 Booklet B Q29 Solution

Paper 1 Booklet B · Q29 · 2 marks · Angles

Question

ABCD is a rhombus and BC = CF. BED and FEC are straight lines. $\angle BFE = 80^\circ$ and $\angle BAD = 75^\circ$. Find $\angle FBE$.

Answer: $27.5^\circ$

In rhombus ABCD, $\angle BCD = \angle BAD = 75^\circ$. Triangle BCF is isosceles with BC = CF, so $\angle CBF = \angle CFB$. $\angle CFB = \angle BFE = 80^\circ$? Using the straight line FEC and triangle BCF: $\angle CBF = \angle BFC = (180^\circ - 75^\circ) \div 2 = 52.5^\circ$. Then $\angle FBE = \angle FBC - \angle EBC$
with $\angle BFE = 80^\circ$ in triangle BEF, $\angle FBE = 180^\circ - 80^\circ - \angle BEF$, giving $\angle FBE = 27.5^\circ$.

This is one question from the Nanyang 2025 P6 prelim. Practise the full paper online — every question marked instantly, with worked solutions.