Question

$PQ$ is a tangent at a point $C$ to a circle with centre $O.$ if $AB$ is a diameter and $\angle CAB={30}^{\circ }$, find $\angle PCA$.

Answer 1

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

To find: $\angle PCA=?$

In $\Delta AOC$,

$\angle CAB=\angle OCA$ (Angles opposite to equal sides are equal)

So, $\angle CAB={30}^o=\angle OCA$

Since $OC\perp PQ$ (Tangent is perpendicular to the radius at point of contact)

$\angle PCO={90}^o$

$\angle OCA+\angle PCA={90}^o$

${30}^o+\angle PCA={90}^o$

$\angle PCA={90}^o-{30}^o$

Therefore, $\angle PCA={60}^o$