Thanks Shadow,
The rest of the question was fine, but they could have made the diagram a bit more specific couldn't they? They had a similar question involving two tangents to a circle that intersected at a point such that the shaded area between the two points the tangents joined at the circle, A and B, was equal to the rest of the area of the circle. That is where I have seen 2*pi - (remaining angle) before.
What I am trying to illustrate is though Shadow, is the f***ing diagram "LOOKED" like it was the radius of the circle, but whatever.
Sorry to bother you again mate, but since my geometry is so bad, could you help me with another one?
I have attached the question and my little sketch. It basically involves two isosceles triangles, one inscribed in another, and I really would like to see how the length BD = 1 + sin(18)
I got all the angles right, but I cannot seem to get the magnitudes of the sides right required. I know BE = 1, then CD = sin(18) and DE = sin(18) from my diagram.
So the length BD should be 1 - 2sin18
If I used the triangle BAD, letting the angle BAD be 54 degrees, then BD = sin(54) = sin(18 + 36) = sin(18)cos(36) + cos(18)sin(36)
Finally, letting angle ABD be 36 degrees, and using BA as the hypotenuse, then BD = cos36 = 1 - 2sin^2(18)
So you can see why I am at a loss?