Question

Question 3
ABCD is a rhombus such that $\angle ACB={40}^{\circ }$, then $\angle ADB$ is
(a) ${40}^{\circ }$
(b) ${45}^{\circ }$
(c) ${50}^{\circ }$
(d) ${60}^{\circ }$

Answer 1

Given, ABCD is a rhombus such that $\angle ACB={40}^{\circ }{\Rightarrow }^{\circ }OCB={40}^{\circ }$

Since, AD ||BC

$\angle DAC=\angle BCA={40}^{\circ }$ [alternate interior angles]

Also, $\angle AOD={90}^{\circ }$ [diagonals of a rhombus are perpendicular to each other]

We know that, sum of all angles of a triangle ADO is ${180}^{\circ }$
$\therefore \angle ADO+\angle DOA+\angle OAD={180}^{\circ }$
$\therefore \angle ADO={180}^{\circ }-({40}^{\circ }+{90}^{\circ })$
$={180}^{\circ }-{130}^{\circ }={50}^{\circ }$
$\Rightarrow \angle ADB={50}^{\circ }$