# Circle question (triangles too)

1. Mar 15, 2007

### Couperin

1. The problem statement, all variables and given/known data

O is the centre of the circle of radius 6cm, and A and B are two points on the circumference such that angle AOB = $$\theta$$ radians.

Show that the length AB is equal to $$\sqrt {72 (1 - cos \theta)}$$

2. Relevant equations

I think the following might be relevant:

Cosine rule: $$\theta = cos^-1\frac{b^2 + c^2 - a^2}{2bc}$$

Area of sector = $$\frac{1}{2}r^2\theta$$

3. The attempt at a solution

I don't really know where to start here. I think pythagoras is involved somewhere. The radius 6 must therefore be squared, multplied by 2 = 72. But I don't want to work backwards from the given soultion.

Am I missing some theory about chords of circles?

Last edited: Mar 15, 2007
2. Mar 15, 2007

### Couperin

No matter, I solved it.

Treat AOB as a triangle. Apply cosine rule to find AB, rearrange and factorise under the squareroot.