Question

$O$ is the centre and PB is a diameter of the circle.BC is the tangent of the circle with B as point of contact.If $\angle APB={60}^0$ then $\angle PBA=?$

Answer 1

We have,

$\angle APB={60}^o$

We know that,

$\angle PAB={90}^o\left(\text{Semicircle}\text{make}\text{angle}\text{at}\text{9}{\text{0}}^{\text{o}}\right)$

Now,

$In\Delta APB$

$\angle A+\angle P+\angle B={180}^o$

$\Rightarrow {90}^o+{60}^o+\angle B={180}^o$

$\Rightarrow \angle B={180}^o-{150}^o$

$\Rightarrow \angle B={30}^o$

Hence, this is the answer.