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$^{\circ}$
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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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