Can you please help me prove that the ratio between the sines of the angles in a triangle is equal to the ratio between the sides of the same triangle:
$$\frac{\sin A}{\sin B \cdot \sin C} = \frac{a}{b \cdot c}$$
Where side a for example is the side opposite to $\angle A$ and same for sides $b$ and $c$.
This equation was introduced to me in class when we where trying to solve a problem:
In $\bigtriangleup$$ABC$, $\sin A:\sin B:\sin C = 2:3:4$. What is the size of the greatest angle in $\bigtriangleup$$ABC$?
The solution I was provided by my teacher included the stated equation without a proof. Keep in mind we only learned the sine and the cosine laws.
Equating two expressions for the area of a triangle gives$$\frac12bc\sin A=\frac12ca\sin B\implies\frac{a}{\sin A}=\frac{b}{\sin B}.$$(In fact, we can prove this ratio is twice the triangle's circumradius.)