Question

Measure of opposite angle of a parallelogram are ${(3x-2)}^o$ and ${(50-x)}^o$. Find the measure of its each angle.

Answer 1

The opposite angles of a parallelogram are given as ${\left(3x-2\right)}^{\circ }$ and ${\left(50-x\right)}^{\circ }$.

Since, the opposite angles of a parallelogram are equal, then,

$3x-2=50-x$

$3x+x=50+2$

$4x=52$

$x=13$

${\left(3x-2\right)}^{\circ }={\left(3\left(13\right)-2\right)}^{\circ }$

$={37}^{\circ }$

Therefore, one angle is ${37}^{\circ }$, and its opposite angle is also ${37}^{\circ }$.

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

$Secondangle={180}^{\circ }-{37}^{\circ }$

$={143}^{\circ }$

Thus, the angles are, ${37}^{\circ }$, ${143}^{\circ }$, ${37}^{\circ }$ and ${143}^{\circ }$.