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In the given figure, PC is a tangent to the circle at C. AOB is the diameter which when extended meets the tangent at P. Find $\angle CBA$ and $\angle BCO$, if $\angle PCA = 110^{\circ}$.
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Sol. $\angle ACB = \angle OCB + \angle OCA = 90^{\circ}$
$\angle PCB + \angle OCB + \angle OCA = 110^{\circ}$
$\angle PCB = 110^{\circ} - 90^{\circ} = 20^{\circ}$
$\angle PCB + \angle OCB = 90^{\circ}$
$\angle OCB = 90^{\circ} - 20^{\circ} = 70^{\circ}$
As $OB = OC \Rightarrow \angle OBC = \angle OCB$
$\angle OBC = \angle OCB = 70^{\circ}$
$\angle PCB + \angle OCB + \angle OCA = 110^{\circ}$
$\angle PCB = 110^{\circ} - 90^{\circ} = 20^{\circ}$
$\angle PCB + \angle OCB = 90^{\circ}$
$\angle OCB = 90^{\circ} - 20^{\circ} = 70^{\circ}$
As $OB = OC \Rightarrow \angle OBC = \angle OCB$
$\angle OBC = \angle OCB = 70^{\circ}$