Question

In the $QP$ and $QR$ are tangents to the circle centre $O$ at $P$ and $R$ respectivlely . Find the value of $x.$

• A. 45
• B. 35
• C. 65
• D. 55

Answer 1

The correct option is A 45

We know that radius are perpendicular to the tangent.

$\therefore$ $\angle ORQ=\angle OPQ={90}^o$

Now,

$\Rightarrow$ $\angle ORQ+\angle RQP+\angle OPQ+\angle POR={360}^o$

$\Rightarrow$ ${90}^o+{50}^o+{90}^o+\angle POR={360}^o$

$\Rightarrow$ ${230}^o+\angle POR={360}^o$

$\therefore$ $\angle POR={130}^o$

$\Rightarrow$ $\angle PSR=\frac{\angle POR}{2}=\frac{{130}^o}{2}={65}^o$

Join $S$ and $O$ the, we get,

$\Rightarrow$ $OR=OS$ [ Radius of a circle ]

$\Rightarrow$ $\angle SRO=\angle OSR={20}^o$ [ Base angles of an equal sides are also equal ]

$\Rightarrow$ $\angle OSP={65}^o-{20}^o={45}^o$

$\Rightarrow$ $\angle SPO=\angle OSP={45}^o$ [ Base angles of an equal sides are also equal ]

$\Rightarrow$ $\angle OPT={90}^o$ [ Radius are perpendicular to the tangent ]

$\Rightarrow$ $\angle SPT={90}^o-{45}^o$

$\Rightarrow$ $\angle SPT={45}^o$

$\therefore$ $x^o={45}^o$