Question

The opposite angles of a parallelogram are $(3x-2{)}^{\circ }$ and $(x+48{)}^{\circ }$. Find the measure of each angle of the parallelogram.

Answer 1

Let four angles of given parallelogram taken in order be $\angle A,\angle B,\angle C$ and $\angle D$ respectively
Here opposite angles given $\angle A=(3x-2)$ and $\angle C=(x+48)$
We know that opposite angles of a parallelogram are equal
$$\therefore 3x-2=x+48$$
$$\begin{array}{cc}& 2x=50\\ & x=25\\ & \therefore 3x-2={73}^{\circ }\\ & \angle A=\angle C={73}^{\circ }\end{array}$$
Also we know that sum of two adjacent angles of a parallelogram is 180
$$\therefore \angle A+\angle B={180}^{\circ }$$
$${73}^{\circ }+\angle B={180}^{\circ }$$
$\angle B={107}^{\circ }=\angle D$ (Opposite angles are equal)
$\therefore$ All four angles of given parallelogram are ${73}^{\circ },{107}^{\circ },{73}^{\circ },{107}^{\circ }$