Question

Find the measure of $\angle ADC$ in the given figure.

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

The correct option is C

${60}^{\circ }$

In $△ABD$,
$\angle DAB+\angle BDA+\angle ABD={180}^{\circ }$
(Sum of all the interior angles of a triangle is ${180}^{\circ }$)

$\Rightarrow \angle BDA={180}^{\circ }-\angle DAB-\angle ABD$$\Rightarrow \angle BDA={180}^{\circ }-{60}^{\circ }-{90}^{\circ }$$\Rightarrow \angle BDA={30}^{\circ }$

Since BC = CD, $△BCD$ is an isosceles triangle.
$\therefore$ $\angle CBD=\angle CDB$
(Angles opposite to equal sides are equal)

In $△BCD$,
$\angle BCD+\angle CBD+\angle CDB={180}^{\circ }$
(Sum of all the interior angles of a triangle is ${180}^{\circ }$)

$\Rightarrow \angle BCD+2\times \angle CDB={180}^{\circ }$

$\Rightarrow {120}^{\circ }+2\times \angle CDB={180}^{\circ }$

$\Rightarrow \angle CDB$ = $\frac{{180}^{\circ }-{120}^{\circ }}{2}$

$\Rightarrow \angle CDB={30}^{\circ }$

From the figure, we know that,
$\angle ADC=\angle ADB+\angle CDB$

$\Rightarrow \angle ADC={30}^{\circ }+{30}^{\circ }$

$\Rightarrow \angle ADC={60}^{\circ }$