Maths Help
posted by Mathslover Please help on .
A,B,C,D,E,F are 6 consecutive points on the circumference of a circle such that AB=BC=CD=10,DE=EF=FA=22. If the radius of the circle is √n, what is the value of n?

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+n2ncosx
2ncosx = 2n100
cosx = (n50)/n
similarly for the larger triangle:
2ncosy = 2n 484
cosy = (n242)/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
n50, hypotenuse = n and height of h
By Pythagoras:
h^2 + (n50)^2 = n^2
h^2 = n^2  (n50)^2 = 100n  2500
h = √(100n2500) = 10√(n25)
sinx = 10√(n25)/n
in the same way:
h2 = 22√(n121)
siny = 22√(n121)/n
cos(x+y) = cosx cosy  sinx siny
cos(120°) = (n50)/n * (n242)/n  (10√(n25))/n * 22√(n121))/n)
1/2 = (n50)/n * (n242)/n  (10√(n25))/n * 22√(n121))/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=%28n50%29%2Fn+*+%28n242%29%2Fn++%2810√%28n25%29%29%2Fn+*+22√%28n121%29%29%2Fn%29+%3D+1%2F2
Check:
cosx = (n50)/n = 218/268
x = 35.567°
cosy = (n121)/n = 26/268
y = 84.433°
x+y = 35.3567 + 84.433 = 120° !!!!!! YEahhh