Question

In a trapezium ABCD, side AB is parallel to side DC. If $\angle A=x^{\circ }$ and $\angle D=(3x-20{)}^{\circ }$;
Find the value of $x$.

Answer 1

As we know that, sum of Co-inerior angles is equal to ${180}^{\circ }$ hence,
$\angle A+\angle D={180}^{\circ }$ --------(1)
$\angle A+\angle D={180}^{\circ }$
$\Rightarrow x+3x-20={180}^{\circ }$
$\Rightarrow 4x={200}^{\circ }$
$\therefore x={50}^{\circ }$

Hence, the value of $x$ is ${50}^{\circ }$