In the figure, $\;$ $PQ = QR$, $\;$ $\angle RQP = 68^\circ$, $\;$ $PC$ and $CQ$ are tangents to the circle with center $O$.
- $\angle QOP$
- $\angle QCP$
Given:
$PQ = QR$, $\;$ $\angle RQP = 68^\circ$, $\;$ $PC, \; CQ$ are tangents, $\;$ $O $ is the center of the circle
- In $\triangle PQR$, $\;$ $\angle PRQ = \angle RQP = 68^\circ$ $\;\;\;$ [$\because \;$ $PQ = QR$, angles opposite equal sides are equal]
$\angle QOP = 2 \angle PRQ$ $\;\;\;$ [angle at the center is twice the angle at remaining circumference]
$\therefore \;$ $\angle QOP = 2 \times 68^\circ = 136^\circ$ - $OP = OQ$ $\;\;\;$ [radii of the same circle]
$\angle OPC = \angle OQC = 90^\circ$ $\;\;\;$ [angle between radius and tangent is $90^\circ$]
$POQC$ is a quadrilateral.
$\therefore \;$ $\angle QOP + \angle OPC + \angle OQC + \angle QCP = 360^\circ$ $\;\;\;$ [sum of angles of a quadrilateral]
$\begin{aligned} \therefore \; \angle QCP & = 360^\circ - \left(\angle QOP + \angle OPC + \angle OQC\right) \\\\ & = 360^\circ - \left(136^\circ + 90^\circ + 90^\circ\right) \\\\ & = 44^\circ \end{aligned}$