Question

The adjacent figure ABCD is parallelogram
$\angle A=3\left(x-5\right)$, $\angle B=\frac{x}{4}$, $\angle C=2\left(x+22\frac{1}{2}\right)$ then show that $\angle A+\angle B={180}^o$

Answer 1

In a parallelogram ABCD, opposite angles are equals. $\angle A=\angle C\Rightarrow 3(x-5)=2(x+22\frac{1}{2})$

$\Rightarrow 3x-15=2x+45\Rightarrow x={60}^o$

$\therefore \angle A=3({60}^o-5)=3\times {55}^o={165}^o$ and $\angle B=\frac{{60}^o}{4}={15}^o$

Now, $\angle A+\angle B={180}^o$

$\Rightarrow {165}^o+{15}^o={180}^o\Rightarrow {180}^o={180}^o$

Hence, this is true.