Question

From the figure given below the tangent at a point $C$ of a circle and a diameter $AB$ when extended intersect at $P$. If $\angle PCA={110}^o$ find $\angle CBA$.

Answer 1

Join C and O.

$\angle BCA={90}^{\circ }$ (angle in semi-circle $={90}^{\circ }$)

Also,$\angle PCO={90}^{\circ }\left(radius\perp tangent\right)$

$\begin{array}{c}\therefore \angle PCA=\angle PCO+OCA\\ \therefore \angle OCA={110}^{\circ }-{90}^{\circ }={20}^{\circ }\\ In\Delta AOC;AO=OC\left(radius\right)\\ \therefore \angle OCA=\angle OAC={20}^{\circ }\\ In\Delta ABC,\\ \angle SBA={180}^{\circ }-\left(\angle BCA+\angle CAB\right)\\ ={180}^{\circ }-\left({90}^{\circ }+{20}^{\circ }\right)={70}^{\circ }\end{array}$