Question

In the above figure, $PC$ is the tangent and $BC$ is the diameter of circle where $O$ is centre of the circle. A secant from point $P$ is drawn that cut the circle at point $A$ and $B$ such that $AB$ is equal to radius of the circle. find $\angle P$.

• A. ${30}^o$
• B. ${60}^o$
• C. ${45}^o$
• D. ${35}^o$

Answer 1

The correct option is A ${30}^o$


In order to find out the value of $\angle P$, we have to construct $OA$ where $O$ is the centre of the circle.

Now, in $△AOB$
$OA=OB=AB\text{(given that AB is equal to radius of circle)}$
So,$△AOB$ is Equilateral triangle,
and hence, $\angle AOB=\angle OBA=\angle OAB={60}^o$

Hence, Using interior angles sum property in $△BPC$,
$\angle BCP+\angle CPB+\angle CBP={180}^o$
${90}^o+{60}^o+\angle CPA={180}^o$
$\angle CPA={180}^o-{150}^o$
$\angle CPA={30}^o$

Hence, $\angle P={30}^o$
So, option $\left(a\right)$ correct.