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:
$$\forall (a, b, c) \in \hspace{0.05em} \mathbb{R}^3, \enspace (\alpha, \beta, \gamma) \in \hspace{0.05em} \mathbb{R}^3, $$
$$ \frac{sin(\alpha)}{a} = \frac{sin(\beta)}{b} = \frac{sin(\gamma)}{c} $$
To show it, let us project a height \( h_c \) upon the length \( c \), and such as the following figure:
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,
$$\forall (a, b, c) \in \hspace{0.05em} \mathbb{R}^3, \enspace (\alpha, \beta, \gamma) \in \hspace{0.05em} \mathbb{R}^3, $$
$$ \frac{sin(\alpha)}{a} = \frac{sin(\beta)}{b} = \frac{sin(\gamma)}{c} $$
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