Question

$ABC$ is an isosceles triangle in the given circle with centre $O,$ if $\angle ABC={42}^{\circ },$ then find the measure of $\angle CDE.$

• A. ${84}^{\circ }$
• B. ${138}^{\circ }$
• C. ${96}^{\circ }$
• D. ${148}^{\circ }$

Answer 1

The correct option is C ${96}^{\circ }$

In $△ABC,$

$AB=AC$ [Given]

We know that angles opposite to equal sides are equal in a triangle. So,

$\angle ACB=\angle ABC={42}^{\circ }$

Now, by using angle sum property,

$\begin{array}{cc}\angle ACB+\angle ABC+\angle A& ={180}^{\circ }\\ {42}^{\circ }+{42}^{\circ }+\angle A& ={180}^{\circ }\\ {84}^{\circ }+\angle A& ={180}^{\circ }\\ \angle A& ={96}^{\circ }\end{array}$

Quadrilateral $ABCD$ is a cyclic quadrilateral so, by using the property that the exterior angle of a cyclic quadrilateral is equal to the opposite interior angle we get,

$\angle CDE=\angle A$

$x={96}^{\circ }$