Question

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

Answer 1

Let $x^{\circ }$ be the smallest angle of the parallelogram.

According to the question, the largest angle is $\left(2x^{\circ }-{24}^{\circ }\right)$.

It is known that the sum of the adjacent angles of the parallelogram is ${180}^{\circ }$. Then,

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

$3x^{\circ }={180}^{\circ }+{24}^{\circ }$

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

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

Therefore, the largest angle is $\left(2\left({68}^{\circ }\right)-{24}^{\circ }\right)={112}^{\circ }$.