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प्रश्न
ABCDE is a regular pentagon. The bisector of angle A meets the side CD at M. Find ∠AMC.
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उत्तर
Given, A pentagoan ABCDE.
The line segment AM is the bisector of the ∠A.
Now, Since the measure of each interior angle of a regular pentagon is 108°.
∴ ∠BAM = `1/2 xx 108^circ` = 54°
By the angle sum property of a quadrilateral, we have (in quadrilateral ABCM)
∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°
⇒ 54° + 108° + 108° + ∠AMC = 360°
⇒ ∠AMC = 360° – 270°
⇒ ∠AMC = 90°
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संबंधित प्रश्न
Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)
| Figure |  |
 |
 |
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| Side | 3 | 4 | 5 | 6 |
| Angle sum | 180° |
2 × 180° = (4 − 2) × 180° |
3 × 180° = (5 − 2) × 180° |
4 × 180° = (6 − 2) × 180° |
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