Question

In the pentagon ABCDE, $\angle A={120}^{\circ },\angle B={138}^{\circ },\angle D=\angle E$. The sides AB and DC, when produced, meet at right angle. Calculate $\angle E$

• A. ${75}^o$
• B. ${65}^o$
• C. ${55}^o$
• D. none of the above

Answer 1

The correct option is A ${75}^o$

Since
$ABP$ is a straight line,
$\angle PBC={180}^o-{138}^o={42}^o$
Since
sum of angles in a triangle BCP $={180}^o$
${90}^o+{42}^o+\angle PCB={180}^o$
$\Rightarrow \angle PCB={48}^o$
Since
$PD$ is a straight line,
$\angle BCD={180}^o-{48}^o={132}^o$
Let $\angle AED=\angle EDC=x$
Since, sum of interior angles in a pentagon is ${540}^o$,

${120}^o+{138}^o+{132}^o+x+x={360}^o$
$\Rightarrow x={75}^o$

Hence, $\angle$E $={75}^o$.