Question

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

Answer 1

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

$2t+7t={180}^{\circ }$
because $\angle DCBand\angle ABC$ are in ratio 2:7]
$t={20}^{\circ }$
$\angle DCB=2t=2\times {20}^{\circ }={40}^{\circ }$

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

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

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

${40}^{\circ }+\angle ADB+\angle ABD={180}^{\circ }$
$\angle ADB+\angle ABD={140}^{\circ }$