Question

ABCDE is regular pentagon and the bisector of $\angle BAE$ meets CD at M . If the bisector of $\angle$ meets AM at P. find $\angle CPA$

• A. ${134}^o$
• B. ${74}^o$
• C. $\angle CPA={144}^{\circ }$
• D. ${154}^o$

Answer 1

The correct option is C $\angle CPA={144}^{\circ }$

Given,

$ABCD$ is a regular pentagon

so eac angle=${108}^{\circ }$

As $AM$ and $CP$ are bisector of angle $BAE$ and $BCD$

$⟹\angle BAP=\angle BCP={25}^{\circ }$

In quad.$ABCP$

$\angle ABP+\angle BAP+\angle BCP+\angle APC=$

$⟹108+54+54+\angle CPA=360$

$⟹\angle CPA=360-216={144}^{\circ }$

Hence $\angle CPA={144}^{\circ }$