Quadrilaterals — Maths STD 9 — Question

 
Given $ABCD$ is a rhombus. Diagonals bisect each other perpendicularly.
Hence $\angle\text{BOC} = 90^\circ$
Given $\angle\text{OCB} = 40^\circ$
$\text{AD || BC}$ and BD is the transversal
$∴ \angle\text{ADB} = \angle\text{DBC}$ (Alternate angles)
Hence in right angled $\triangle\text{BOC},$
$\angle\text{BOC} + \angle\text{OCB} + \angle\text{OBC} = 180^\circ$
$⇒ 90^\circ + 40^\circ + \angle\text{OBC} = 180^\circ$
$⇒130^\circ + \angle\text{OBC} = 180^\circ$
$⇒ \angle\text{OBC} = 180^\circ - 130^\circ = 50^\circ$
But $\angle\text{OBC} = \angle\text{DBC}$
Therefore, $\angle\text{ADB} = 50^\circ.$