In fig. $3$ , $P Q$ and $P R$ are tangents to the circle with centre $O$ and $S$ is a point on the circle such that $∠ S Q L = 50^{\circ}$ and $∠ S R M = 60^{\circ} .$ Find $∠ Q S R .$

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Solution

Join $O Q$ and $O R$
$∠ O Q P = ∠ O R P = 90^{\circ}$
$⟹ O Q S = 90^{\circ} - 50^{\circ} = 40^{\circ}$
$⟹ O R S = 90^{\circ} - 60^{\circ} = 30^{\circ}$
In $△ O Q S, O S = O Q = r a d i u s ⟹ ∠ O S Q = ∠ O Q S = 40^{\circ}$
In $△ O R S, O S = O R = r a d i u s ⟹ ∠ O S R = ∠ O R S = 30^{\circ}$
$⟹ ∠ Q S R = ∠ O S R + ∠ O S Q = 30^{\circ} + 40^{\circ} = 70^{\circ}$

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