Question

In a pentagon $ABCDE$, $BC$ is parallel to $ED$ and $\angle B:\angle A:\angle E=5:3:4$. Find $\angle B$.

• A. ${130}^{\circ }$
• B. ${150}^{\circ }$
• C. ${170}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is B ${150}^{\circ }$

In a pentagon ABCDE, BC is parallel to ED and $\angle B:\angle A:\angle E=5:3:4$
$\angle C+\angle D={180}^o$ (Sum of interior angles of the same sides of the transversal are supplementary)
Sum of total interior angles of pentagon = ${180}^o(n-2)={180}^o(5-2)={540}^o$
Let $\angle B=5x,\angle A=3x,\angle E=4x$
Now, $\angle C+\angle D+\angle B+\angle A+\angle E={540}^o$
$=>{180}^o+5x+3x+4x={540}^o$
$=>12x={360}^o$
$=>x={30}^o$
$\angle B=5x=5\times {30}^o={150}^o$