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December 20, 2014

December 20, 2014

Posted by **Mathslover Please help** on Wednesday, May 8, 2013 at 1:53am.

- Maths Help -
**Reiny**, Wednesday, May 8, 2013 at 7:41amWe have 3 isosceles triangles with a base of 10 and

3 isosceles triangles with a base of 22

let the central angle of each of the smaller be x and the angle of each of the larger be y

3x + 3y = 360°

x+y = 120°

for the smaller using the cosine law

10^2 = √n^2 + √n^2 - 2√n √n cosx

100 = n+n-2ncosx

2ncosx = 2n-100

cosx = (n-50)/n

similarly for the larger triangle:

2ncosy = 2n- 484

cosy = (n-242)/n

cos(x+y) = cosx cosy - sinx siny

so we need sinx and sin y

Make a sketch of a right -angled triangle with base

n-50, hypotenuse = n and height of h

By Pythagoras:

h^2 + (n-50)^2 = n^2

h^2 = n^2 - (n-50)^2 = 100n - 2500

h = √(100n-2500) = 10√(n-25)

sinx = 10√(n-25)/n

in the same way:

h2 = 22√(n-121)

siny = 22√(n-121)/n

cos(x+y) = cosx cosy - sinx siny

cos(120°) = (n-50)/n * (n-242)/n - (10√(n-25))/n * 22√(n-121))/n)

-1/2 = (n-50)/n * (n-242)/n - (10√(n-25))/n * 22√(n-121))/n)

What a horrible equation:

So I relied on Wolfram to do all the drudgery and amazingly got

**n = 268**

http://www.wolframalpha.com/input/?i=%28n-50%29%2Fn+*+%28n-242%29%2Fn+-+%2810√%28n-25%29%29%2Fn+*+22√%28n-121%29%29%2Fn%29+%3D+-1%2F2

Check:

cosx = (n-50)/n = 218/268

x = 35.567°

cosy = (n-121)/n = 26/268

y = 84.433°

x+y = 35.3567 + 84.433 = 120° !!!!!! YEahhh

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