Question

Solve : $O$ is the centre and $PB$ is a diameter of the circle, $PA\perp BA$. $BC$ is the tangent of the circle with $B$ as point of contact. If $\angle APB={60}^{\circ }$ then
( i ) $\angle PBC=?$
(ii) $\angle ABC=?$
(iii) $\angle PBA=?$
(iv) $\angle ABD=?$

Answer 1

$\begin{array}{c}\left(i\right)\angle PBC={90}^0\left(ii\right)\left(anglebetweentangentandradiusis{90}^o\right)\\ ii)\angle ABC={90}^0-{30}^0={60}^0\\ \left(iii\right)\angle PBA={30}^0\\ \left(iv\right)\angle ABD={90}^0+{30}^0={120}^0\end{array}$