ABCDE is a regular pentagon. The bisector of angle A meets the aide CD at M. Find angle AMC
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See, I have circumscribed the regular Pentagon with a circle. Since, the 360° of the circle is divided equally by the pentagon to five parts, i.e., 72°, the angle COD =72°.
Now, I have calculated angle CAD=36°, which is half of angle COD, using the property of a circle.
Now, I have calculated angle ACD, using property of triangle, in ∆ACD.
36°+ angle ACD+ angle ADC= 180
Since, ∆ACD is an isosceles ∆,
36° +2× angle ACD= 180°
angle ACD = 72°
Now finally take a look at the ∆AMC.
Angle MAC = half of angle CAD= 18°
Now,
angle MAC+angle AMC+angle ACM=180°
18°+angle AMC + 72° = 180°
angle AMC = 90°
HENCE PROVED!
TRY TO USE THE PROPERTIES, DIRECTLY OR INDIRECTLY. PRACTICE MORE! GUD LUCK!
Now, I have calculated angle CAD=36°, which is half of angle COD, using the property of a circle.
Now, I have calculated angle ACD, using property of triangle, in ∆ACD.
36°+ angle ACD+ angle ADC= 180
Since, ∆ACD is an isosceles ∆,
36° +2× angle ACD= 180°
angle ACD = 72°
Now finally take a look at the ∆AMC.
Angle MAC = half of angle CAD= 18°
Now,
angle MAC+angle AMC+angle ACM=180°
18°+angle AMC + 72° = 180°
angle AMC = 90°
HENCE PROVED!
TRY TO USE THE PROPERTIES, DIRECTLY OR INDIRECTLY. PRACTICE MORE! GUD LUCK!
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