Question

In a parallelogram, ABCD, find the measure of all the angles if one of its angles is ${15}^{\circ }$ less then twice the smallest angle.

Answer 1

In parallelogram $ABCD,$

Let $\angle A$ be $x$. Then, $\angle B$ will be $2x-{15}^o$

We know that, in a parallelogram sum of adjacent angles are supplementary.

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

$\Rightarrow$ $x+2x-{15}^o={180}^o$

$\Rightarrow$ $3x={195}^o$

$\Rightarrow$ $x={65}^o.$

$\therefore$ $\angle A={65}^o$

$\Rightarrow$ $\angle B=2x-{15}^o=2\times {65}^o-{15}^o={115}^o$.

We know that, in parallelogram opposite angles are equal.

$\therefore$ $\angle A=\angle C$ and $\angle B=\angle D$

$\therefore$ $\angle C={65}^o$ and $\angle D={115}^o.$

$\therefore$ The measure of all angles of a parallelogram are ${65}^o,{115}^o,{65}^o$ and ${115}^o.$