Question

In the diagram, PT and QT are tangents to the circle, centre O, at P and Q respectively. Find the value of $x$.

• A. ${30}^o$
• B. ${40}^o$
• C. ${50}^o$
• D. ${60}^o$

Answer 1

The correct option is B ${40}^o$

$\angle QOP={140}^0$
$\angle QUP=\frac{140}{2}={70}^0$
$\Rightarrow \angle QRP=180-70={110}^0$ .....[QRPU is a cyclic quadrilateral]
$\angle OQR+\angle QRP+\angle RPO+\angle POQ={360}^0$ ...[QRPO is a quadrilateral]
$\angle OPR=360-[140+110+60]$
$=360-310$
$={50}^0$
$\angle OPR+\angle RPT={90}^0[OP\perp PT]$
$X^0=\angle RPT=90-{50}^0={40}^0$