For any triangles with vertex angles and corresponding opposite sides are A, B, C and a, b, c, respectively, the sine law is given by the formula...

**Derivation**

To derive the formula, erect an altitude through B and label it h_{B} as shown below. Expressing h_{B} in terms of the side and the sine of the angle will lead to the formula of the sine law.

$\sin A = \dfrac{h_B}{c}$

$h_B = c \sin A$

$\sin C = \dfrac{h_B}{a}$

$h_B = a \sin C$

Equate the two h_{B}'s above:

$h_B = h_B$

$c \sin A = a \sin C$

$\dfrac{c}{\sin C} = \dfrac{a}{\sin A}$

To include angle B and side b in the above relationship, construct an altitude through C and label it h_{C} as shown below.

$\sin A = \dfrac{h_C}{b}$

$h_C = b \sin A$

$\sin B = \dfrac{h_C}{a}$

$h_C = a \sin B$

$h_C = h_C$

$b \sin A = a \sin B$

$\dfrac{b}{\sin B} = \dfrac{a}{\sin A}$

Thus,

Therefore, the ratio of one side to the sine of its opposite angle is constant.

**Note:**

The constant ratio above is the diameter of the circumscribing circle about the triangle. See the proof (*not available for now*) for this note.