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Nan Hua 2025 P6 Maths Prelim — Paper 1 Booklet A Q12 Solution

Paper 1 Booklet A · Q12 · 2 marks · Angles

Question

In the figure, $BCD$ is an equilateral triangle. $ABD$ is an isosceles triangle with $AD = BD$ and $\angle ABD = 46^\circ$. Find $\angle DCA$.

1. $16^\circ$
2. $37^\circ$
3. $44^\circ$
4. $46^\circ$

Answer: 1

Worked solution

\(BCD \text{ equilateral: } BD = DC, \angle BDC = 60^\circ\)
\(ABD \text{ isosceles with } AD = BD, \angle ABD = \angle BAD = 46^\circ\)
\(\angle ADB = 180^\circ - 2 \times 46^\circ = 88^\circ\)
\(\angle ADC = \angle ADB + \angle BDC = 88^\circ + 60^\circ = 148^\circ\)
\(AD = BD = DC, \text{ so } \triangle ACD \text{ isosceles}\)
\(\angle DCA = (180^\circ - 148^\circ) \div 2 = 16^\circ\)

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