ABCDE is a regular polygon the bisector of angle of the pentagon meets the side CD at m show that angle A andAngleC is equal to 90degree
2. the sum of the exterior angles of polygon is1/9 of the sum of interior angle . find the no. of side of this polygon
2. the sum of the exterior angles of polygon is1/9 of the sum of interior angle . find the no. of side of this polygon
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Answer:
1)→ 90°
Step-by-step explanation:
Each interior angle of a regular pentagon =
n
2n−4
×90
=
5
2×5−4
×90
0
=6×18=108
0
∴∠BAM=
2
1
(108
0
)=54
0
In the quadrilateral ABCM,
∠BAM+∠ABC+∠BCM+∠AMC=360
0
⇒54
0
+108
0
+108
0
+∠AMC=360
0
⇒∠AMC=90
0
(Refer figure in hint if needed).