SOLUTION: - find the number of sides in a polygon whose interior angles are in arithmetical progression. the smallest being 120° and the common difference 5 °
Question 983960: - find the number of sides in a polygon whose interior angles are in arithmetical progression. the smallest being 120° and the common difference 5 ° Found 3 solutions by ikleyn, Alan3354, MathTherapy:Answer by ikleyn(52805) (Show Source): You can put this solution on YOUR website!
The sum of the interior angles of a given polygon is the sum of arithmetic progression with the first term 120° and the common difference of 5°, according to the condition.
So, it is
degrees.
From the other side, it is 180°*(n-2), according to the formula of sum of interior angles of a polygon.
Thus you have an equation
= .
Simplify and solve it step by step:
= ,
= ,
= ,
= ,
= .
This quadratic equation has two roots: n = 9 and n = 16 (use the quadratic formula or the Vieta's theorem).
So, the problem has two solutions: n = 9 and n = 16.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website! I worked this one.
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The other tutor gave answers of 9 and 16.
Both satisfy the equation, but 16 sides gives one angle that is 180 degs, then some that are greater than 180.
An angle of 180 doesn't count as an angle.
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--> only the 9 sides solution works.
120 + 125 + 130 + 135 + 140 + 145 + 150 + 155 + 160 = 1260 --> 9 sides
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for 16 sides,
+ 165 + 170 + 175 + 180 ... That's NG. Answer by MathTherapy(10553) (Show Source): You can put this solution on YOUR website!
- find the number of sides in a polygon whose interior angles are in arithmetical progression. the smallest being 120° and the common difference 5 °