Question

If one angle of parallelogram is ${36}^{\circ }$ less than twice itsadjacent angle then, find the angles of a parallelogram.

Answer 1

Let $ABCD$ is a parallelogram. Let $\angle ADC=x$.

Therefore,

$\angle DCB=2x-{36}^{\circ }$ [Given that one angle is ${36}^{\circ }$ less than twice the adjacent angle.]

Now, we know that adjacent angles of a parallelogram are supplementary.

$\Rightarrow \angle ADC+\angle DCB={180}^{\circ }$

$\Rightarrow x+2x-{36}^{\circ }={180}^{\circ }$

$\Rightarrow 3x={216}^{\circ }$

$\Rightarrow x=\frac{{216}^{\circ }}{3}=2={72}^{\circ }$

Hence, $\angle ADC={72}^{\circ }$ and $\angle DCB=\times {72}^{\circ }-{36}^{\circ }={108}^{\circ }$

Also, opposite angles of a parallelogram are equal ,therefore,

$\Rightarrow \angle ABC=\angle ADC={72}^{\circ }$ and $\angle BAD=\angle DCB={108}^{\circ }$

Thus, the four angles of the parallelogram are ${108}^{\circ },{72}^{\circ },{108}^{\circ },{72}^{\circ }$.