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In the given figure, PQ and PR are tangents to the circle with centre $O$ and $S$ is a point on the circle such that $\angle{SQL}={50}^{\circ}$ and $\angle{SRM}={60}^{\circ}$. Find $\angle{QSR}.$

What I've tried,

Join $OQ$ and $OR$. Since the line joining the point of contact of the tangent to the centre of the circle is equal to $90^{\circ}$.

$\therefore$$\angle$OQL=$\angle{ORL}={90}^{\circ}$

But, now I am stuck.

enter image description here

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  • $\begingroup$ I think you mean $\angle OQL=90^\circ$. $\endgroup$ – Gregory Grant Nov 25 '15 at 16:54
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Draw your two radii. $\triangle SOR$ is isoceles and $\angle SRO=30^\circ=\angle RSO$. Similarly $\angle SQO=40^\circ=\angle QSO$, so $\angle QSR=70^\circ$

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Since $\angle QRS = \angle LQS = 50$ and $\angle RQS = \angle SRM = 60$, we have $\angle QSR = 180-\angle QRS-\angle RQS = 180-50-60=\boxed{70^\circ}$

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  • $\begingroup$ How can you take QRS as an angle? $\endgroup$ – Abhishekstudent Nov 25 '15 at 17:05
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The big tools you need with this specific problem is that:

1) all radiuses are equal

2) isosceles triangle theorem (angles opposite... are equal)

3) Tangents to the circle make a 90 degree angle with the segment from point of intersection and the center of the circle.

Take note that

$$OS=OR$$

$$OS=OQ$$

Draw the tangents and add the angles $30$ and $40$ which I see you've erased not so throughly. Also take note of any isosceles triangles. Then I hope the problem will be obvious to you.

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