Question 13

In the figure, PQ is a diameter of the circle. Angle PQS = $$35^\circ$$. Find angle QRS.

Solution

SInce PQ is the diameter, the angle subtended by it at R is 90 deg. i.e., $$\angle$$ PRQ = 90 deg.

Let $$\angle$$ RPQ = $$\theta$$, then $$\angle$$RQP = 90 - $$\theta$$

As the angles subtended by a chord in same segment are equal, $$\angle$$RPQ = $$\angle$$RSQ = $$\theta$$

In triangle RSQ, $$\angle$$QRS + $$\angle$$RSQ + $$\angle$$RQS = 180 

$$\Rightarrow$$ $$\angle$$QRS + $$\theta$$ + 35 + 90 - $$\theta$$ = 180

$$\Rightarrow$$ $$\angle$$QRS = 180 - 125 = 55 deg.

Hence $$\angle$$QRS = 55 deg.

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