The angles of a hexagon are in arithmetic sequence. Prove that its smallest angle is always greater than 60 degree
Answers
Given : smallest angle is greater than 60 degree
To find : prove that its smallest angle is greater than 60 degree
Solution:
Sum of angle of polygon of n sided = (n- 2) * 180°
Hexagon has 6 sides so
Sum of all angles = (6 - 2) * 180° = 720°
Let say smallest angle = a° a > 0
and d° is the common difference
then largest angle = a + 5d
largest angle should be less than 180°
=> a + 5d < 180°
=> a + 5d = 180 - k k > 0
Sum of all angles
a + a + d + a + 2d + a + 3d + a + 4d + a + 5d = 720
=> 6a + 15d = 720
=> 2a + 5d = 240
=> a + a + 5d = 240
=> a + 180 - k = 240
=> a = 60 + k
=> a > 60
QED
Hence proved
smallest angle is greater than 60°
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