Question
In a triangle PQR, ∠P = 60° and ∠Q = ∠R, find ∠R.
✓
Answer
Let $\angle Q=\angle R=x, \angle P=60^{\circ}$
But $\angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180^{\circ}$
$\therefore 60^{\circ}+\mathrm{x}+\mathrm{x}=180^{\circ}$
$\Rightarrow 60^{\circ}+2 \mathrm{x}=180^{\circ}$
$\Rightarrow 2 \mathrm{x}=180^{\circ}-60^{\circ}=120^{\circ}$
$\Rightarrow \mathrm{x}=\frac{120^{\circ}}{2}=60^{\circ}$
$\therefore \angle \mathrm{Q}=\angle \mathrm{R}=60^{\circ}$
Hence, $\angle \mathrm{R}=60^{\circ}$
But $\angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180^{\circ}$
$\therefore 60^{\circ}+\mathrm{x}+\mathrm{x}=180^{\circ}$
$\Rightarrow 60^{\circ}+2 \mathrm{x}=180^{\circ}$
$\Rightarrow 2 \mathrm{x}=180^{\circ}-60^{\circ}=120^{\circ}$
$\Rightarrow \mathrm{x}=\frac{120^{\circ}}{2}=60^{\circ}$
$\therefore \angle \mathrm{Q}=\angle \mathrm{R}=60^{\circ}$
Hence, $\angle \mathrm{R}=60^{\circ}$
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