Question

Given: PQ is a tangent. AB is a diameter, $\angle CAB={30}^{\circ }\angle PCA=?$

Answer 1

In $\Delta AOC,$
As, AO =CO (Given)
So, $\angle CAO=\angle OCA$ (Angles opposite to equal sides are equal)
or $\angle CAB=\angle OCA$
but $\angle CAB={30}^{\circ }$
So, $\angle OCA={30}^{\circ }...\left(i\right)$
Since, $OC\perp PQ$ (Tangent is perpendicular to the radius at point of contact)
$\Rightarrow \angle PCO={90}^{\circ }\Rightarrow \angle OCA+\angle PCA={90}^{\circ }\Rightarrow {30}^{\circ }+\angle PCA={90}^{\circ }\Rightarrow \angle PCA={90}^{\circ }-{30}^{\circ }\therefore \angle PCA={60}^{\circ }$