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