Question

ABCDE is a pentagon in which AB is parallel to ED. If $\angle B={142}^{\circ },\angle C=3x^{\circ }$ and $\angle D=2x^{\circ }$. Calculate $x$.

• A. ${110}^{\circ }$
• B. ${160}^{\circ }$
• C. ${130}^{\circ }$
• D. ${\left(43\frac{3}{5}\right)}^o$

Answer 1

The correct option is D ${\left(43\frac{3}{5}\right)}^o$

In a pentagon ABCDE, AB is parallel to ED and $\angle B={142}^{\circ },\angle C=3x^{\circ }$ and $\angle D=2x^{\circ }$
So, $\angle A+\angle E={180}^o$ (Sum of interior angles on the same side of transversal are supplementary.)
Sum of all interior angles of pentagon = ${180}^o(n-2)={180}^o(5-2)={540}^o$
Now,
$\angle A+\angle E+\angle B+\angle C+\angle D={540}^o$
$\Rightarrow {180}^o+{142}^o+3x+2x={540}^o$
$\Rightarrow 5x={218}^o$
$\Rightarrow x={\left(43\frac{3}{5}\right)}^o$