Question : If $\triangle \mathrm{BPQ} \cong \triangle \mathrm{ASR}$ and $\angle \mathrm{A}=\frac{1}{3} \angle \mathrm{R}=\angle \mathrm{S}$, then find $\angle \mathrm{Q}$. (All angles are in degrees).
Option 1: 108°
Option 2: 36°
Option 3: 72°
Option 4: 118°
Correct Answer: 108°
Solution :
Given, $\angle \mathrm{A}=\frac{1}{3} \angle \mathrm{R}=\angle \mathrm{S}$
⇒ $3\angle \mathrm{A}=\angle \mathrm{R}$ and $\angle \mathrm{A}=\angle \mathrm{S}$
In $\triangle \mathrm{ASR},$ $\angle A + \angle S +\angle R = 180°$
⇒ $\angle \mathrm{A} + \angle \mathrm{A}+3\angle \mathrm{A}=180°$
⇒ $5\angle \mathrm{A} = 180°$
⇒ $\angle \mathrm{A} = \frac{180°}{5}=36°$
Now, $\triangle \mathrm{BPQ} \cong \triangle \mathrm{ASR}$
⇒ $\angle \mathrm{Q}=\angle \mathrm{R}$
⇒ $\angle \mathrm{Q}=3\angle \mathrm{A}$
⇒ $\angle \mathrm{Q}=3\times 36° = 108°$
Hence, the correct answer is 108°.
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