Question

$ABCDE$ is a regular pentagon. The bisector of angle $A$ of the pentagon meets the side $CD$ in point $M$. Find the measure of $\angle AMC$.

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

The correct option is D ${90}^o$

Sum of interior and exterior of a pentagon is ${540}^{\circ }$ & each angle is ${108}^{\circ }$.

$S=(n-2)\times {180}^{\circ }$

$=(5-2)\times {180}^{\circ }={540}^{\circ }$

Each interior angle $\frac{{540}^{\circ }}{5}={108}^{\circ }$

In quad. $ABCM$

$\angle MAB+\angle B+\angle C+\angle AMC={360}^{\circ }$

AM is bisector of $\angle A$ , therefore $\angle MAB=\frac{108}{2}={54}^{\circ }$

${54}^{\circ }+{108}^{\circ }+{108}^{\circ }+\angle AMC={360}^{\circ }$

$\angle AMC={360}^{\circ }-{270}^{\circ }={90}^{\circ }$