Catholic High 2025 P6 Maths Prelim — Paper 1 Booklet A Q11 Solution

Paper 1 Booklet A · Q11 · 2 marks · Angles

Question

In the figure, FCEB and DGE are straight lines. ABC is an equilateral triangle. $DGE = FCE$ and $\angle CFD = 70^\circ$. Find $\angle CGE$.

Answer: 2

\triangle DEF \text{ has } DE = FE, \text{ so } \angle EDF = \angle EFD = 70^\circ
\angle DEF = 180^\circ - 70^\circ - 70^\circ = 40^\circ
\angle GEC = \angle DEF = 40^\circ
\angle GCE = \angle ACB = 60^\circ \text{ (equilateral triangle)}
\angle CGE = 180^\circ - 40^\circ - 60^\circ = 80^\circ

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