Quadrilaterals — Maths STD 9 — Question

3. 50º

Solution:

ABCD is a rhombus such thet $\angle\text{ACB}=40^\circ.$

We know that diagonnals of rhombus bisect each other right angles.

In right $\Delta\text{BOC},$ we have

$\angle\text{OBC}=180^\circ-(\angle\text{BOC}+\angle\text{BCO})$ (angle sum property)

$=180^\circ-(90^\circ+40^\circ)=50^\circ$

$\therefore\ \angle\text{DBC}=\angle\text{OBC}=50^\circ$

Now,

$\angle\text{ADB}=\angle\text{DBC}$ [Alt. int. $\angle\text{s}$ ]

$\therefore\ \angle\text{ADB}=50^\circ[\therefore\angle\text{DBC}=50^\circ]$

Hence, (c) is the correct answer.