Question

In figure, ABCD is a parallelogram. If $\angle DAB={60}^{\circ }and\angle DBC={80}^{\circ },\angle CDB$ is :

• A. ${40}^{\circ }$
• B. ${80}^{\circ }$
• C. ${60}^{\circ }$
• D. ${20}^{\circ }$

Answer 1

The correct option is A ${40}^{\circ }$

Given $ABCD$ is a parallelogram and $BD$ is the diagonal.

$AB\parallel CD$

$\Rightarrow \angle CBD=\angle ADB={80}^o$

We have $\angle DAB=\angle BCD={60}^o$

Diagonal $BD$ forms a triangle $CBD,$

$\Rightarrow \angle DCB+\angle CBD+\angle CDB={180}^o$ [ Sum of $\angle$ les of a triangle }

$\Rightarrow {60}^o+{80}^o+\angle CDB={180}^o$

$\therefore \angle CBD={40}^o$ $,$ $\angle ADB={80}^o$

Hence, the answer is ${40}^o.$