Why is the diameter of a circumcircle of a triangle related to the law of sines? Please refer to the following image for clarity. 
In this diagram I must find AC ( which is$\sqrt21$) in order to find the radius of the circle. I know that the radius can be found by dividing the diameter by $2$, and I know that the diameter is equal to AC/sin(60). My question is, why is this true? Why is the diameter equal to this?


Let $ABC$ a triangle with acute angle $\gamma$, let $M$ be the center of the circumcircle and $r$ be its radius.  By moving $C$ on the arc over $AB$ the angle $\gamma$ doesn't change; move it to the intersection $C'$ of $AM$ and the circle.  From Thales the triangle $ABC'$ is right-angled with a right angle at $B$.  Hence $\sin(\gamma)=AB/2r$.
