Question

Find the measure of each angle of a $parallelogram$, one its angle is $30$ less than twice the smaller angle.

Answer 1

Let measure of smaller angle $A$ be $x$, then measure of major angle $B$ be $(2x-30)$

$\Rightarrow$ $\angle A+\angle B={180}^o$ [ Sum of adjacent angles in parallelogram is supplementary ]

$\Rightarrow$ $x+(2x-30)={180}^o$

$\Rightarrow$ $3x-30={180}^o$

$\Rightarrow$ $3x={210}^o$

$\Rightarrow$ $x={70}^o$

$\Rightarrow$ $\angle A=x={70}^o$

$\Rightarrow$ $\angle B=(2x-30{)}^o=(2\times 70-30{)}^o={110}^o$

We know that opposite angles are equal in parallelogram

$\therefore$ $\angle A=\angle C={70}^o$

$\therefore$ $\angle B=\angle D={110}^o$