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

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

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