Question : In $\triangle P Q R, S$ is a point on the side QR such that $\angle Q P S=\frac{1}{2} \angle P S R, \angle Q P R=78^{\circ}$ and $\angle P R S=44^{\circ}$. What is the measure of $\angle PSQ$?
Option 1: 68$^{\circ}$
Option 2: 56$^{\circ}$
Option 3: 58$^{\circ}$
Option 4: 64$^{\circ}$
Correct Answer: 64$^{\circ}$
Solution :
$\angle Q P S=\frac{1}{2} \angle P S R$
Let $\angle PSR = 2\theta$ and $\angle QPS = \theta$
Sum of all the sides of the triangle = $180^{\circ}$
⇒ $\angle QPR + \angle PRS + \angle RQP = 180^{\circ}$
⇒ $78^{\circ} + 44^{\circ} + \angle RQP = 180^{\circ}$
⇒ $122^{\circ} + \angle RQP = 180^{\circ}$
⇒ $ \angle RQP = 180^{\circ}- 122^{\circ} = 58^{\circ}$
$\angle PSR$ = exterior angle of $\angle PSQ$
⇒ $2\theta = \theta + 58^{\circ}$
⇒ $2\theta - \theta = 58^{\circ}$
⇒ $\theta = 58^{\circ}$
$\angle PSQ = 180^{\circ} - 2\theta$
$= 180^{\circ} - 2×58^{\circ}$
$= 180^{\circ} - 116^{\circ}$
$= 64^{\circ}$
Hence, the correct answer is 64$^{\circ}$.
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