Question

Find the measure of all the angles of a parallelogram, if one angle is ${24}^{\circ }$ less than twice the smallest angle.

Answer 1

Given: One angle is ${24}^{\circ }$ less than twice the smallest angle

Let $ABCD$ is a parallelogram and $\angle D=x$ be the smallest angle.

$\therefore$ Other angle $\angle C=2x-{24}^{\circ }$

We know that the adjacent angles of a paralleloagram is supplementary.

$\therefore x+2x-{24}^{\circ }={180}^{\circ }$

$\Rightarrow 3x={204}^{\circ }$

$\Rightarrow x={68}^{\circ }$

$\therefore \angle C=2x-{24}^{\circ }=2\times {68}^{\circ }-{24}^{\circ }={112}^{\circ }$

$\because$ Opposite angles of parallelogram are equal
$\Rightarrow$ The measure of $\angle A\text{and}\angle B\text{are}{112}^{\circ }\text{and}{68}^{\circ }$ respectively.

Hence, the angles of the parallelogram are ${68}^{\circ },{112}^{\circ },{68}^{\circ }\text{and}{112}^{\circ }$ respectively.