Question

In fig. 1, 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, $\angle CAB={30}^{\circ }$and PQ is a tangent at a point C to a circle with centre O.

Since, AB is a diameter.

$\therefore \angle ACB={90}^{\circ }$

Join OC

$\angle CAO=\angle ACO={30}^{\circ }(OA=OC)$

and, $\angle PCO={90}^{\circ }$ (Tangent is perpendicular to the radius through the point of contact)

$\therefore \angle PCA=\angle PCO-\angle ACO$

= ${90}^{\circ }-{30}^{\circ }={60}^{\circ }$