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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Angles of a polygon.
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nadeemoidu (1189)

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The angles of a polygon are in AP with common difference 5 degrees .
The smallest angle is 120 degree. Find the no. of sides in the polygon.
    
tarinbansal (3903)

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Arre isme kya hai?
The formula of the sum of angles of any polygon is = 180(n-2) where n is the no of sides. Now the no. of sides=no of angles.
Hence equate the sum of angles by the 2 formulas- AP's sum and the formula I gave you.

n/2[240 + (n-1)5] = 180(n-2)

You will get the value of n.

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nadeemoidu (1189)

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There is some thing. Find the final answer and I'll let u know
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chetan_kp (283)

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clearly no. of sides = no. of angles
largest angle possible = 175
using AP formula
tn = a + ( n - 1 ) X d
175 = 120 + ( n - 1 ) X 5
n - 1 = 11
hence n = 12
no.of sides = 12

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tarinbansal (3903)

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I am getting n=16,9

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tarinbansal (3903)

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Chetan_kp,
ur logic that no angle in a polygon can be greater than 180 is wrong.
look at the figure, it is a polygon and angle A in more than 180.


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tarinbansal (3903)

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Wat's the trick U were talking about?

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nadeemoidu (1189)

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A polygon cannot have an angle equal to 180 degree.
 
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tarinbansal (3903)

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Oh yes. U R right. So the answer shud be either 15 or 8. Isn't it?

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tarinbansal (3903)

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Its a X class question and I forgot this point while solving it.
Silly me.

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nadeemoidu (1189)

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@ tarin How did u get 15, 8 ? Its not ur silly mistake , its a tricky question.

The answer is 9 only . If it has 16 sides  , then the angles are 120, 125,.....180, 185, 190, 195 . But a polygon cannot have an angle equal to 180.


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tarinbansal (3903)

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Oh ya.
Again a silly mistake. Oh god.
Actually I did this question when I was in X.
I think I must think a little more before solving these type of questions.

The quality of a person's life is in direct proportion to their commitment to excellence, regardless of their chosen field of endeavor.

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chimanshu_007 (11437)

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Hello
 
let the no. of sides = n , sum of interior angles =
 
Sn = (2n-4).90
 
first term (angle , a) = 120o , d = 5o
 
Sn = (n/2) [2.120 + (n-1).5]
 
so
 
(2n-4).90 = (n/2)[240 + 5(n-1)]
 
(2n-4).180 = n(5n + 535)
 
solve this....u will get an =n
 
n2 -25n + 144 = 0
 
(n-16)(n-9) = 0
 
n = 16 OR 9
 
if n = 16....the last angle , a + (n-1)d = 120 + 15.5 = 195
 
its nt possible...
 
so n = 9
 
HOPE THIS HELPS :)

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