In the figure, $PQ$ and $PR$ are tangents to a circle with center $O$.
Given: $\;$ $\angle QPR = 60^\circ$; $\;$ $PQ, \; PR =$ tangents to the given circle
$O$ is the center of the given circle
$\angle OQP = \angle ORP = 90^\circ$ $\;\;\;$ [angle between radius and tangent]
In quadrilateral $OQPR$,
$\angle QOR + \angle ORP + \angle OQP + \angle APR = 360^\circ$ $\;\;\;$ [sum of angles of a quadrilateral]
$\therefore \;$ $\angle QOR = 360^\circ - \left(\angle ORP + \angle OQP + \angle QPR\right)$
i.e. $\;$ $\angle QOR = 360^\circ - \left(90^\circ + 90^\circ + 60^\circ\right) = 120^\circ$
In $\triangle OQR$,
$OQ = OR$ $\;\;\;$ [radii of the same circle are equal]
$\therefore \;$ $\angle OQR = \angle ORQ$ $\;\;\;$ [angles opposite equal sides are equal]
$\angle OQR + \angle ORQ + \angle QOR = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]
i.e. $\;$ $2 \angle OQR + \angle QOR = 180^\circ$
i.e. $\;$ $2 \angle OQR = 180^\circ - \angle QOR = 180^\circ - 120^\circ = 60^\circ$
$\implies$ $\angle OQR = 30^\circ$
Now, $\;$ $\angle QSR = \dfrac{1}{2} \angle QOR$ $\;\;\;$ [angle at center is twice the angle at remaining circumference]
i.e. $\;$ $\angle QSR = \dfrac{120^\circ}{2} = 60^\circ$