Question

ABCD is a rhombus. If $\angle ACB={40}^o$, then $\angle ADB$ is

• A. ${90}^o$
• B. ${80}^o$
• C. ${60}^o$
• D. ${50}^o$

Answer 1

Answer: (D)

In given figure ABCD is a rhombus. We know that diagonals of rhombus bisect each other perpendicularly.

Hence, $\angle$BOC= 90${}^{\circ }$

$\angle$OCB = 40${}^{\circ }$ (Given)

AD$\parallel$BC and BD is the transversal --- (Opposite sides of rhombus are parallel to each other)

$\therefore$ $\angle$ADB = $\angle$DBC ---- (Alternate angles)

In $△$OBC,

$\angle$BOC + $\angle$OCB + $\angle$OBC = 180${}^{\circ }$

$\Rightarrow$ 90${}^{\circ }$+ 40${}^{\circ }$ + $\angle$OBC = 180${}^{\circ }$

$\Rightarrow$ $\angle$OBC =180${}^{\circ }$ - 130${}^{\circ }$

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

But $\angle$OBC =$\angle$DBC

$\therefore$ $\angle$ADB = 50${}^{\circ }$ ---( Alternate angle)