Question

In a pentagon $ABCDE,AB$ is parallel to $DC$ and $\angle A:\angle E,\angle D=3:4:5$. Find $\angle E$ degree.

Answer 1

$\angle B+\angle C={180}^o$ ($\because AB||CD$ Sum of alll interior angle are same side of transversal is ${180}^o$)

$\begin{array}{c}\angle A=3x\\ \angle E=4x\\ \angle D=5x\end{array}\}x=$ constant of proportionality

We know, Sum of interior angle of pentagon $={540}^o$

or, $\angle A+\angle B+\angle C+\angle D+\angle E={540}^O$

OR, $5x+{180}^o+4z+5z={540}^o$

or, $12x={360}^o$

$\therefore x={30}^o$

$\therefore \angle E=4\times {30}^o={120}^o$