Question

ABCD is rhombus such that $\angle ACB={40}^{\circ }$ then $\angle ADC$ :

• A. ${40}^{\circ }$
• B. ${45}^{\circ }$
• C. ${100}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

Answer: (C)

${100}^{\circ }$

Given ABCD is a rhombus

The diagonals bisect each other perpendicularly.

$\Rightarrow \angle BOC=90°$ $\angle OCB=40°$

$AB\parallel BC$ and $BD$ is transversal.

$\Rightarrow \angle BDA+\angle OCB+\angle OBC=180°$

$\therefore {90}^o+{40}^o+\angle OBC=180°$

$\therefore \angle OBC=50°$

$\Rightarrow \angle OBC=\angle DBC=50°$

$\therefore \angle ADB=50°$

$\therefore \angle ADC=50°+50°=100°$

Hence, the answer is $100°.$