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The interior angles of a polygon are in ...

The interior angles of a polygon are in arithmetic progression. The smallest angle is `120^@` and the common difference is `5^@` Find the number of sides of the polygon

Text Solution

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Let there be n sides in the polygon. Then by geometry, the sum of alln interior angles of polygon is `(n-2)xx180^(@)`. Also, the angles are in A.P. with the smallest angle `120^(@)` and common difference `5^(@)`.
Therefore. Sum of all interior angles of polygon is
`n/2[2xx120+(n-1)xx5]`
Thus, we must have
`n/2[2xx120+(n-1)5]=(n-2)xx180`
or `n/2[5n+235]=(n-2)xx180`
or `5n^(2)+235n=360n-720`
or `5n^(2)-125n+720=0`
or `n^(2)-25n+144=0`
or (n-16)(n-9)=0
or n=16,9
But if n=16, then `16^(th)` angle=`120+15xx5=195gt180^(@)` which is not possible. Hence, n=9.
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