Question

In pentagon $ABCDE$, sides $AB$ and $AE$ are perpendicular to each other, $\angle B={112}^{\circ }$, $AB$ and $DC$ when produced meet at right angle and $\angle D:\angle E=1:2$. Find $\angle E$ (in degrees)

• A. ${140}^{\circ }$
• B. ${120}^{\circ }$
• C. ${130}^{\circ }$
• D. none of the above

Answer 1

The correct option is B ${120}^{\circ }$

Given :

$\angle BFC=90°$

$\angle A=90°$

$\angle B=112°$

Solution:

As $\angle BFC=\angle BAE=90°$

Therefore, $FD\text{is parallel to}AE$

Hence, $\angle D+\angle E=180°$ … (angles on the same side of transversal are supplementary)

Since $\angle E=2\times \angle D$

Therefore, $\frac{\angle E}{2}+\angle E=180°$

$\frac{3\times \angle E}{2}=180°$

$\angle E=120°$