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The difference between any two consecutive interior angles of a polygon is 5°.If the smallest angle is 120°, find the number of the sides of the polygon.

The solution is simple..the formula for finding the sum of interior angles of a n sided polygon =180 degrees * (n-2..

therefore using this we get

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The formula to find the sum of interior angles of an n sided polygon is given by 180 degrees * (n-2)

the smallest angle is given as 120 and angle between two sides is given as 5 degrees

therefore the angles will form an A.P. with a=120 and d=5

from this we get

 

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Yeah but you will have to reject 16 because the 16th term = a + (n-1) d = 120 + (15)5 = 120 + 75 = 195

And as we can't have an angle on a polynomial with an angle greater than 180, we have to reject the idea of such a 16 sided polygon because the greatest angle will exceed 180 degrees.

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but non convex polygon's angle can exceed 180 right ??
fr eg - star....it has interior angles which is greater than 180

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The answer has to be 9. ... not 16 bcoz...if the polygon had 16 sides it would hv....195° as its last angle(i dntknow if THIS is possible or not).....but acc. to this...the 13th angle would be 180°…which clearly makes this condition insensible.. :)
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Aishwarya.... from where did u get this question??? Can u tell me pleaSe!!
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okay,n=9 and also n=16 i know n=16 is not possible as angle of polygon should not exceed 180 degree but what is the meaning of n=16 as maths did not give answer without any reason??
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This question is from NCERT 11 N=16 is not valid as The exterior angle when n=16 is 60+15×-5 Which is -15 The sum of an AP is quadratic eqn It has two roots and often one is invalid as ^2 increases one root
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By Daksh

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But why is 195° for any quadrilateral invalid; if the quadrilateral is convex then?
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If cosec theta + cot theta=p

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