Question

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

Answer 1

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

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

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

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

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

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