Question

In the figure given below, find z.

• A. ${40}^{\circ }$
• B. ${110}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is C ${45}^{\circ }$

$\angle ABC={180}^{\circ }–{110}^{\circ }={70}^{\circ }$ [linear pair]

AB = AD [given]

$\Rightarrow \angle ABD=\angle ADB={70}^{\circ }$ [angles opposite to equal sides are equal]

$\angle ADC={180}^{\circ }–\angle ADB={180}^{\circ }–{70}^{\circ }=110°$

In $△DAC$
$\angle DAC+\angle ADC+\angle ACD={180}^{\circ }$

$\angle DAC={180}^{\circ }–\angle ADC–\angle ACD$

$\therefore \angle DAC={180}^{\circ }–{110}^{\circ }–{25}^{\circ }={45}^{\circ }$
$\therefore z={45}^{\circ }$