#### 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 Figure].

Solution
Given : $\angle PCA= 110^{\circ}$
Here $\angle PCA= 90^{\circ}$
[$\because$  PC is tangent]
$\angle PCA= \angle PCO+\angle OCA$
$110^{\circ}= 90^{\circ}+\angle OCA$
$\angle OCA= 20^{\circ}$
$\therefore \: \angle OCA= \angle OAC= 20^{\circ}\cdots \left ( i \right )$ [$\because$  Sides opposite to equal angles are equal]
We know that PC is tangent and angles in alternate segment are equal.
Hence $\angle BCP= \angle CAB= 20^{\circ}$
In $\bigtriangleup OAC$
$\angle O+\angle C+\angle A= 180^{\circ}$ [Interior angles sum of triangle is 180°]
$\angle O+20^{\circ}+20^{\circ}= 180^{\circ}$ [using (i)]
$\angle O= 180^{\circ}-40^{\circ}= 140^{\circ}$  …..(ii)
Here $\angle COB+\angle COA= 180^{\circ}$

$\angle COB= 180^{\circ}-140^{\circ}$( using (ii))
$\angle COB= 40^{\circ}$ …..(iii)
In $\bigtriangleup COB$
$\angle C+\angle O+\angle B= 180^{\circ}$ [using (iii)]
$90^{\circ}-20^{\circ}+40^{\circ}+\angle B= 180^{\circ}$
$\angle B= 180^{\circ}-110^{\circ}= 70^{\circ}$
Hence $\angle CBA= 70^{\circ}$