Question 92

In the given figure, PQ is the diameter of the circle. What is the measure (in degrees) of ∠QSR?

Solution

$$\angle$$ QSR = $$\angle$$ QPR = $$\theta$$ [Angles in the same segment]

Also, $$\angle$$ PRQ = $$90^\circ$$ [Angle in the semi circle]

In $$\triangle$$ PQR, => $$\angle$$ PQR + $$\angle$$ PRQ + $$\angle$$ QPR = $$180^\circ$$

=> $$43^\circ+90^\circ+\theta=180^\circ$$

=> $$\theta=180-133=47^\circ$$

=> Ans - (C)

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