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OG Geometry - Find the external angles of the 5 pointed star

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OG Geometry - Find the external angles of the 5 pointed star

by Lasve » Sun Jan 08, 2012 6:37 am
Hi Guys I'm writing about a tricky question that took me quite a while to solve.

I wanted to ask your opinion about the solve method given by the OG, because I find it quite confusing and time consuming.

I think that a faster and easyer way to solve it would be:
i. Realize the center is a pentagon
ii. That means that the total internal angles are 180*(n-2) = 180 * (5-2) = 540
iii. That means that the Average internal angle is 108
iv. Realize that each internal angle is part of a 180-degrees Straight angle, That means that the complementary one (the Base of the triangle) is 180-108= 72
v. Since every triangle is 180 degree, the external angle must be 180-(72*2) = 36
vi. Multiply 36 for the number of external angles: 36*5=180

What do you think?
Did I get something wrong and solve it just for sheer luck or is it ok?
Thanks![/img]
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by LalaB » Sun Jan 08, 2012 11:15 am
hm, have doubts about step III (u divide 540 by 5 ,assuming that all angles are equal. r they?)

my approach-

a+b+c+d+e=180(5-2)=540
remember-the measure of an exterior angle equals the sum of its two non-adjacent interior angles

lets assume that the triangle with the angle z has also angles m (near c) and p(near b)
then c=p+z b=m+z c+b=p+m+2z=180+z
same with the rest angles of the polygon abcde.

now pay attention to the fact that the angles of the polygon counted twice (so we need to divide the result by 2). having this in the mind, we got the following-

(180+z+180+v+180+w+180+x+180+y)/2=540

5*180+(z+w+v+x+y)=540*2
z+w+v+x+y=180
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by rijul007 » Sun Jan 08, 2012 11:40 am
Lasve wrote:Hi Guys I'm writing about a tricky question that took me quite a while to solve.

I wanted to ask your opinion about the solve method given by the OG, because I find it quite confusing and time consuming.

I think that a faster and easyer way to solve it would be:
i. Realize the center is a pentagon
ii. That means that the total internal angles are 180*(n-2) = 180 * (5-2) = 540
iii. That means that the Average internal angle is 108
iv. Realize that each internal angle is part of a 180-degrees Straight angle, That means that the complementary one (the Base of the triangle) is 180-108= 72
v. Since every triangle is 180 degree, the external angle must be 180-(72*2) = 36
vi. Multiply 36 for the number of external angles: 36*5=180

What do you think?
Did I get something wrong and solve it just for sheer luck or is it ok?
Thanks![/img]
This approach would have been correct if the pentagon was regualar, but we dont know that.
You got the correct answer because you were asked the sum of all angles v,w,x,y,z.
If the ques would have been, find x+y+z.
you would have got 36*3 = 108, which is wrong.
the data would be insufficient fot that summation.


check ou the figure
Image

a = v+m
e = v+n
a+e = v+m+n+v = 180+v
similarly similarly you can get expressions for other pairs of adjacent angles
a+b+b+c+c+d+d+e+e+a = 5(180) +v+w+x+y+z
2(a+b+c+d+e) = 900 + v+w+x+y+z
2(540) = 900 + v+w+x+y+z
1080-900 = v+w+x+y+z
v+w+x+y+z = 180

Option C
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by Lasve » Mon Jan 09, 2012 4:32 am
Thank's for the answers!

Anyway, It is obvious that this method would not have worked if I had been asked to find only 3 of the sides!
But in that case it would have been impossible to determine them with no further information.

Moreover I treat the penthagon as regular because, since I'm asked to find out the sum of all angles, lesser and larger angles would compensate each other.
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by GMATGuruNY » Mon Jan 09, 2012 5:07 am
Check here for an efficient approach (similar to that of the original poster, but more visual):

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by ArunangsuSahu » Mon Jan 09, 2012 1:34 pm
180 degree
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by Lasve » Thu Jan 12, 2012 3:44 am
Thank's Guru, I did exactly as in the thread you post!
(althought I took me more than 10 minutes to do it!)
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