Question

In the adjacent figure, tangents drawn through point $A$ and $C$ intersects at point $P$. if $\angle APC={50}^o$, then find $\angle ABC$.

Answer 1

R.E.F. image.

Given : $\angle APC={50}^{\circ }$

To find : $\angle ABC$

Let O be the center of

the circle

const.- Join OA and OC

$\angle OAP=\angle OCP={90}^{\circ }$ [line joining the center with the point of contact

of tangent makes ${90}^{\circ }$ with the tangent]

In quadrilateral AOCP, by Angle sum property

$\angle OAP+\angle APC+\angle PCO+COA={360}^{\circ }$

$\angle COA={130}^{\circ }$

$\angle COA=2\angle BAC$ [Angle subtended by on are at the center is double the

angle subtended by it in any point on the circle ]

$\angle BAC=\frac{\angle COA}{2}=\frac{{130}^{\circ }}{2}$

$\angle BAC={65}^{\circ }$