# ABCDE is a regular pentagon. The bisector of angle A meets the side CD at M. Find ∠AMC.

**Solution:**

Given, ABCDE is a regular __pentagon__.

The bisector of angle A meets the side CD at M.

We have to find ∠AMC.

We know, measure of each interior angle of a polygon = (2n - 4)/n × 90°

Here, n = 5

Each interior angle = (2(5) - 4)/5 × 90°

= 6/5 × 90°

= 6 × 18°

= 108°

The measure of each interior angle of a regular pentagon is 108°.

Since AM bisects the angle A,

∠BAM = 1/2 × 108° = 54°

We know that the sum of all the angles of a __quadrilateral__ is 360°.

In quadrilateral ABCM,

∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°

54° + 108° + 108° + ∠AMC = 360°

∠AMC = 360° - 270°

Therefore, ∠AMC = 90°

**✦ Try This: **A diagonal and a side of a rhombus are of equal length. Find the measure of the angles of the rhombus.

**☛ Also Check: **NCERT Solutions for Class 8 Maths

**NCERT Exemplar Class 8 Maths Chapter 5 Problem 159**

## ABCDE is a regular pentagon. The bisector of angle A meets the side CD at M. Find ∠AMC.

**Summary:**

ABCDE is a regular pentagon. The bisector of angle A meets the side CD at M. The measure of ∠AMC is 90°.

**☛ Related Questions:**

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