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The tangent at a point C of a circle and a diameter AB when extended intersect at P. If \anglePCA =110^{\circ} , find \angleCBA [see Figure].

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   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}
Here OC = OA (Radius)
\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}

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