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?

 A: If $AB$ and $BC$ are $5$ and $4$, as indicated, then 
$$
AC^2=AB^2+BC^2-2|AB||BC|\cos 60=25+16-40\frac12=21
$$ 
by the law of cosines. Hence $|AC|=\sqrt{21}$, not $\sqrt{7}$. The circumdiameter is the product of all sides divided by twice the area, and the area is half the product of two adjacent sides times the sine of the angle between them:
$$
D=\frac{abc}{2A}=\frac{abc}{2\left(\frac12 ab\sin\angle(a,b)\right)}
=\frac{c}{\sin\angle(a,b)}.
$$
So what's left is the opposite side divided by twice the sine of the angle, i.e. $AC/\sin(60)$. 
This explains why all the ratios in the law of sines are equal, they are all equal to the circumdiameter. Basically, this is because one can create a new right triangle, where the diameter is opposite the right angle, and one other side is a side of the original triangle with the same opposite angle. This is based on the Inscribed Angle theorem, see the linked Wikipedia proof.
A: Add the center of the circumcircle at $O$ and drop the perpendicular onto $\overline{AC}$ from $O$ at $D$.

The Inscribed Angle Theorem says that $\angle AOC$ is twice $\angle ABC$, so that $\angle AOD=\angle COD=\angle ABC$. Therefore,
$$
\overline{AC}=2r\sin(B)
$$
A: 
Simple construction. AE is the diameter through circum-center $O$. $\angle ACE= 90^{\circ},$ subtended in semi-circle $ADCE$. Subtended angles on$B$ side of $AC$ are always the same in magnitude.
So, the hypotenuse or diameter
$$ AE =\frac{AC}{\sin 60^{\circ}}$$
Repeat this thrice, once each for three of the vertices A,B or C and you have already derived the Law of Sines:
$$ \frac{a}{\sin A} = \frac{b}{\sin B}=\frac{c}{\sin C}= AE $$
A: If your question is why $\frac{AD}{\sin 60}= 2R$ check the vídeo 
link
A: 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 $\sin(\gamma)=AB/2r$.
