Question on: WAEC Mathematics - 2009
Each of the interior angles of a regular polygon is 140o. Calculate the sum of all the interior angles of the polygon
A
1080o
B
1260o
C
1800o
D
2160o
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Correct Option: B
Each interior angle = 140
\(\frac{(n - 2) \times 180}{n} = 140\)
(n - 2) x 180 = 140n
150 - 360 = 140n
180m - 140n = 360
40n - 360
n = \(\frac{360}{40}\)
n = 9
Sum of all interior angles = (n - 2) x 180
= (9 - 2) x 180
= 7 x 180
= 1260
\(\frac{(n - 2) \times 180}{n} = 140\)
(n - 2) x 180 = 140n
150 - 360 = 140n
180m - 140n = 360
40n - 360
n = \(\frac{360}{40}\)
n = 9
Sum of all interior angles = (n - 2) x 180
= (9 - 2) x 180
= 7 x 180
= 1260
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