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(C) Let $O$ be the center of the circle. Join $OC$. Since $OC$ is the radius and $PC$ is a tangent at $C,$ $OC \perp PC.$ Therefore,$\angle PCO = 90^{

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(C) Let $O$ be the center of the circle. Join $OC$. Since $OC$ is the radius and $PC$ is a tangent at $C,$ $OC \perp PC.$ Therefore,$\angle PCO = 90^{\circ}.$Given $\angle PCA = 110^{\circ},$ we have $\angle OCA = \angle PCA - \angle PCO = 110^{\circ} - 90^{\circ} = 20^{\circ}.$In $	riangle OCA,$ $OC = OA$ (radii of the same circle),so $\angle OAC = \angle OCA = 20^{\circ}.$Since $\angle COA = 180^{\circ} - (20^{\circ} + 20^{\circ}) = 140^{\circ},$ the angle subtended by arc $AC$ at the center is $140^{\circ}.$The angle subtended by arc $AC$ at the circumference is $\angle ABC = \frac{1}{2} \angle COA = \frac{1}{2} 	imes 140^{\circ} = 70^{\circ}.$Thus,$\angle CBA = 70^{\circ}.$

The tangent at a point $C$ of a circle and a diameter $AB$ when extended intersect at $P$. If $\angle PCA = 110^{\circ},$ find $\angle CBA$ [see $Fig.$]. (in $^{\circ}$)

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