Question

Two opposite angles of a parallelogram are $(3x-2)$ and $(63-2x)$.Find all the angles of the parallelogram.

Answer 1

In a parallelogram opposite angles are equal.
So, $(3x-2)=(63-2x)$
$\Rightarrow 3x-2=63-2x$
$\Rightarrow 3x+2x=63+2$
$\Rightarrow 5x=65$
$x=\frac{65}{5}$
$\therefore x=13$
Substituting $x=13$ in $3x-2=63-2x$,we get
$(3\times 13)-2=63-(2\times 13)$
$\Rightarrow 39-2=63-26$
$\therefore 37=37$
So, the angles are ${37}^{\circ }$, ${180}^{\circ }-{37}^{\circ }$, ${37}^{\circ }$ and ${180}^{\circ }-{37}^{\circ }$
i.e. ${37}^{\circ },{143}^{\circ },{37}^{\circ }$and ${143}^{\circ }$