Question

In the given figure, PQ = QR, $\angle RQP={68}^{\circ }$, PC and CQ are tangents to the circle with center O. Calculate the values of:

$\left(i\right)\angle QOP$

$\left(ii\right)\angle QCP$ [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 3 Marks

$\left(i\right)In\Delta PQR$,

$PQ=QR\Rightarrow \angle PRQ=\angle QPR$

[Angles opp. to equal sides of a D are equal]

Also, $\angle QPR+\angle RQP+\angle PRQ={180}^{\circ }$

$\Rightarrow {68}^{\circ }+2\angle PRQ={180}^{\circ }$

$\Rightarrow 2\angle PRQ={180}^{\circ }-{68}^{\circ }={112}^{\circ }$

$\Rightarrow \angle PRQ={56}^{\circ }$.

$\therefore \angle QOP=2\angle PRQ=(2\times {56}^{\circ })={112}^{\circ }$. [Angle at the center is double the Angle on the circle]

Since the radius through the point of contact is perpendicular to the tangent, we have

$\angle OQC={90}^{\circ }$ and $\angle OPC={90}^{\circ }$.

Now, $\angle OQC+\angle QOP+\angle OPC+\angle QCP={360}^{\circ }$

$\Rightarrow {90}^{\circ }+{112}^{\circ }+{90}^{\circ }+\angle QCP={360}^{\circ }$.

$\Rightarrow \angle QCP=({360}^{\circ }-{292}^{\circ })={68}^{\circ }$.