Question : In $\triangle {PQR} $, PQ = PR and S is a point on QR such that $\angle {PSQ}=96^{\circ}+\angle {QPS}$ and $\angle {QPR} = 132^{\circ}$. What is the measure of $\angle {PSR}$?
Option 1: 45°
Option 2: 56°
Option 3: 54°
Option 4: 52°
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Correct Answer: 54°
Solution :
Let $\angle QPS$ be $\theta$
⇒ $\angle PSQ = 96^\circ + \theta$
Now, In $\triangle PQR$
⇒ $\angle Q + \angle R = 180^\circ - \angle P$
$= 180^\circ -132^\circ$
$=48^\circ$
Since $PQ = PR$
$\angle Q = \angle R$
$\therefore \angle Q = \angle R = 24^\circ$
Now, In $\triangle PQS$
$(96^\circ + \theta ) + \theta + 24^\circ = 180^\circ$
⇒ $2\theta = 60^\circ$
⇒ $\theta = 30^\circ$
⇒ $\angle PSQ = 96^\circ + 30^\circ = 126^\circ $
$\therefore \angle PSR = 180^\circ - 126^\circ=54^\circ$
Hence, the correct answer is $54^\circ$.
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