In the figure at left, PQ is a diameter of a circle with centre O. If $∠ P Q R = 55^{o}, ∠ S P R = 25^{o}$ and $∠ P Q M = 50^{o}$ .
Find (i) $∠ Q P R$ ,
(ii) $∠ Q P M$ and
(iii) $∠ P R S$ .

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Solution

$∠ P R Q = ∠ P M Q = 90^{\circ}$ (angle in the semi-circle)

In $△ P R Q$

(i) $∠ Q P R = 180 - ∠ P R Q - ∠ P Q R = 180 - 90 - 55 = 35^{\circ}$

In $△ P M Q$

(ii) $∠ Q P M = 180 - 50 - 90 = 40^{\circ}$

$∠ S P Q = ∠ S P R + ∠ R P Q = 25 + 35 = 60^{\circ}$

(iii) PQRS is a cyclic quadrilateral. So opposite angles are supplementary.

$⟹ ∠ S P Q + ∠ S R Q = 180^{\circ}$

$∠ S R Q = 180 - 60 = 120^{\circ}$

$∠ S R P + ∠ P R Q = 120^{\circ}$

$⟹ ∠ P R S = 120 - 90 = 30^{\circ}$

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