Question

ABCD is a parallelogram in which $\angle DCB$ and $\angle ABC$ are in ratio 2:7. Find the sum of $\angle ABD$ and $\angle ADB$.

• A. ${40}^{\circ }$
• B. ${50}^{\circ }$
• C. ${70}^{\circ }$
• D. ${140}^{\circ }$

Answer 1

The correct option is D

${140}^{\circ }$

$\angle DCB+\angle ABC={180}^{\circ }$
[Adjacent angles of a parallelogram are supplementary]

$\Rightarrow 2t+7t={180}^{\circ }$
[$\because \angle DCB$ and $\angle ABC$ are in ratio 2:7]

$\Rightarrow t={20}^{\circ }$

$\therefore \angle DCB=2t=2\times {20}^{\circ }={40}^{\circ }$

and $\angle ABC=7t=7\times {20}^{\circ }={140}^{\circ }$

Now, $\angle DAB=\angle DCB={40}^{\circ }$
[Opposite angles of a parallelogram are equal]

$\angle DAB+\angle ADB+\angle ABD={180}^{\circ }$
[Sum of the angles of a triangle]

$\Rightarrow {40}^{\circ }+\angle ADB+\angle ABD={180}^{\circ }$

$\Rightarrow \angle ADB+\angle ABD={140}^{\circ }$