Question

ABCD is cyclic quadrilateral. The tangents to a circle at A and C meet at P. If $\angle$ APC = 50${}^{\circ }$, then the value of $\angle$ ADC is:

• A. ${65}^{\circ }$
• B. ${70}^{\circ }$
• C. ${80}^{\circ }$
• D. none of these

Answer 1

The correct option is A ${65}^{\circ }$

Given $\angle APC={50}^0$

As the radius is always perpendicular to tangent at the point of contact,

$\angle OCP=\angle OAP={90}^0$

As $AOCP$ is a quadrilateral, sum of all its interior angles are equal to ${360}^0$

$\therefore \angle AOC={130}^0[{360}^0-({90}^0+{90}^0+{50}^0\left)\right]$

We know that, angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc on the circumference of the circle.

$\therefore \angle ADC=\frac{{130}^0}{2}$

$={65}^0$

Hence, $\angle ADC={65}^0$.