Question

ABCDE is a regular pentagon and bisector of angleBAE meets CD at M. Of bisector of angle BCD meets AM at P. Find angle CPM

Plz help frnds....​

Answer 1

$\huge\bf{Solution: }$

We know that the measure of each interior angle of regular Pentagon is 108°.

$\therefore \sf \angle \: BAM = \dfrac{1}{2} (108 ^{ \circ} ) = {54}^{o}$

Sum of angles of quadrilateral is 360°. Therefore,In Quadrilateral ABCM, we have :

$\star \: \sf \angle BAM + \angle ABC + \angle BCM + \angle CMA = {360}^{o}$

$\longrightarrow \sf \: {54}^{o} + {108}^{o} + {108}^{o} + \angle CMA = {360}^{o}$

$\longrightarrow \sf \: \angle CMA = {90}^{o} ...(i)$

Since CP is the bisector of angle BCD

$\therefore \sf \angle \: PCM = {54}^{o}$

Now,In Triangle CPM,we have :

$\star \: \sf \angle \: PCM + \angle CMP + \angle CPM = {180}^{o}$

$\longrightarrow \sf \: {54}^{o} + {90}^{o} + \angle CPM = {180}^{o}$

$\longrightarrow \sf \: \angle CPM = \large\boxed{{ \sf \: 36}^{o} }$

Answer 2

$\huge{\boxed{\boxed{\tt{Answer:}}}}$

Given:

ABCDE is a regular pentagon and bisector of angle BAE meets CD at M. Of bisector of angle BCD meets AM at P.

Find:

Find angle CPM.

Solution:

540° = total angle of pentagon.

108° = each angle degree.

Let "y" be the variable.

For quadrilateral ABCP:

$\sf\rightarrow{\angle =54° PABPAB}$

$\sf\rightarrow{\angle=108° ABC}$

$\sf\rightarrow{\angle =54°BCP}$

$\sf\rightarrow{\angle=y° CPA}$

ASM(Angle sum property) of a quadrilateral:

$\sf\rightarrow{\angle \: PAB +\angle ABC + \angle BCP +\angle \: CPA = 360°}$

$\sf\rightarrow{54° + 108° + 54° + x° = 360°}$

$\sf\rightarrow { x° = 144°}$

linear pair:

$\sf\rightarrow{\angleCPM = 180° - y°}$

$\sf\rightarrow{\angle CPM = 36°}$

Therefore, 36° is the angle CPM.