if smallest angle of a Pentagon Is 40°,what is the common difference?
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Step-by-step explanation:
Sum of the interior angles of a polygon of n sides.
=(2n−4)π/2=(n−2)π=(n−2)⋅180
o
.
Also a=120
o
,d=5
o
.
∴
2
n
[2⋅120
o
+(n−1)5
o
]=(n−2)⋅180
o
Cancel 5 and simplify.
∴n
2
−25n+144=0
∴(n−9)(n−16)=0.
∴n=9,16.
But when n=16,T
n
=a+(16−1)d
=120
o
+15×5
o
=195
o
This is not possible as interior angle cannot be greater than 180
o
,
∴n=9 is the correct answer.
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