The law of sines

In the context of an unspecified triangle $\{a, b, c\}$, with each angle in front of its respective length, such as:

$$ \left \{ \begin{gather*} \alpha \enspace opposé \enspace à \enspace a \\ \beta \enspace opposé \enspace à \enspace b \\ \gamma \enspace opposé \enspace à \enspace c \end{gather*} \right \} $$

And such as the following figure:

The law of sines tells us that:

$$\forall (a, b, c) \in \hspace{0.05em} \mathbb{R}^3, \enspace (\alpha, \beta, \gamma) \in \hspace{0.05em} \mathbb{R}^3, $$

$$ \frac{sin(\alpha)}{a} = \frac{sin(\beta)}{b} = \frac{sin(\gamma)}{c} $$

Demonstration

To show it, let us project a height $ h_c $ upon the length $ c $, and such as the following figure:

Projection of an height on one the triangle sides

Straightaway, the following relations come:

$$ sin(\alpha) = \frac{h_c}{b} \qquad (1) $$

$$ sin(\beta) = \frac{h_c}{a} \qquad (2) $$

Dividing the equation $ (1) $ by $ a $, we do have:

$$ \frac{sin(\alpha)}{a} = \frac{h_c}{ab} \qquad (3) $$

In the same way, dividing $ (2) $ by $ b $:

$$ \frac{sin(\beta)}{b} = \frac{h_c}{ab} \qquad (4) $$

We now notice that both right memebers of $ (3) $ and $ (4) $ are equals, it follows that:

$$ \frac{sin(\alpha)}{a} = \frac{sin(\beta)}{b} \qquad (5) $$

By reproducing this operation on the two others lengths, we will have two new equations:

$$ \frac{sin(\beta)}{b} = \frac{sin(\gamma)}{c} \qquad (6) $$

$$ \frac{sin(\gamma)}{c} = \frac{sin(\alpha)}{a} \qquad (7) $$

Equalities $ (5), (6), (7) $ having a common member from one to another, they are all equals.

And finally,

$$\forall (a, b, c) \in \hspace{0.05em} \mathbb{R}^3, \enspace (\alpha, \beta, \gamma) \in \hspace{0.05em} \mathbb{R}^3, $$

$$ \frac{sin(\alpha)}{a} = \frac{sin(\beta)}{b} = \frac{sin(\gamma)}{c} $$