Hello, GreenDay14!

Prove the Law of Sines for triangle ABC: .$\displaystyle \frac{\sin A}{a} \:=\: \frac{\sin B}{b} \:=\: \frac{\sin C}{c}$ Here is the classic textbook proof. Code:

C
*
*| *
* | *
b * | * a
* |h *
* | *
* | *
* | *
A * - - - * - - - - - - - * B
D

Draw altitude $\displaystyle CD$ to side $\displaystyle AB.$

. . Call it $\displaystyle h.$

In right triangle $\displaystyle CDA\!:\;\;\sin A \:=\:\frac{h}{b} \quad\Rightarrow\quad h \:=\:b\sin A$ .[1]

In right triangle $\displaystyle CDB\!:\;\;\sin B \:=\:\frac{h}{a} \quad\Rightarrow\quad h \:=\:a\sin B$ .[2]

Equate [1] and [2]: .$\displaystyle b\sin A \:=\:a\sin B \quad\Rightarrow\quad \boxed{\frac{\sin A}{a} \:=\:\frac{\sin B}{b}}$

In a similar fashion, we can prove that: .$\displaystyle \frac{\sin B}{b} \:=\:\frac{\sin C}{c}$