Question

In the given figure (PQ is diameter), x is equal to

• A. ${30}^o$
• B. ${45}^o$
• C. ${70}^o$
• D. ${60}^o$

Answer 1

The correct option is D ${60}^o$

Join $CR$.

$CR$ is a radius and $AB$ is a tangent.

$CR\perp AB$ [ Radius if perpendicular to the tangent ]

$\therefore$ $\angle ARC={90}^o$

In $△CPR$,

$\Rightarrow$ $CP=CR$ [ Radius of a circle ]

$\Rightarrow$ $\angle CPR=\angle CRP$ [ Base angles of equal sides also an equal ]

$\therefore$ $\angle CRP={30}^o$ [ Since, $\angle CPR={30}^o$ ] --- ( 2 )

$\Rightarrow$ Now $\angle ARC=\angle ARP+\angle CRP$

$\Rightarrow$ ${90}^o=x+{30}^o$ [ From ( 1 ) and ( 2 ) ]

$\Rightarrow$ $x={90}^o-{30}^o$

$\therefore$ $x={60}^o.$