Question

In a parallelogram ABCD if $\angle A$ is $\frac{4}{5}$ of $\angle B$, then what is $\angle A$?

Answer 1

Since from the property of parallelogram,

The sum of angle on the same side of the line $AB$ is equal to

${180}^o$, i.e.

$\angle A+\angle B={180}^o$

$\angle A=\frac{4}{5}\angle B$ (Given)

$\Rightarrow \frac{4}{5}\angle B+\angle B={180}^o$

$=\frac{9}{5}\angle B={180}^o$

$=\angle B=\frac{180\times 5}{9}={180}^o$

$=\angle B={100}^o$

$\Rightarrow \angle A=\frac{4}{5}\angle B=\frac{4}{5}\times 100$

$\angle A={80}^o$

$\angle B={100}^o$