## Task B-3.2.

For the angles $\alpha, \beta$ and $\gamma$ of the acute-angled triangle $ABC$, it holds that $\operatorname{tg} \alpha: \operatorname{tg} \beta: \operatorname{tg} \gamma=1: 3: 12$. Calculate

$$
\sin \alpha \cdot \sin \beta \cdot \sin \gamma
$$
## Solution.

Since the angles of the triangle are acute, their tangents have the same sign, and we have:

$$
\begin{aligned}
& \operatorname{tg} \gamma=\operatorname{tg}[\pi-(\alpha+\beta)]=-\operatorname{tg}(\alpha+\beta)=-\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta} \\
& \operatorname{tg} \gamma-\operatorname{tg} \gamma \cdot \operatorname{tg} \alpha \cdot \operatorname{tg} \beta=-\operatorname{tg} \alpha-\operatorname{tg} \beta \\
& \operatorname{tg} \gamma+\operatorname{tg} \alpha+\operatorname{tg} \beta=\operatorname{tg} \alpha \cdot \operatorname{tg} \beta \cdot \operatorname{tg} \gamma
\end{aligned}
$$

Since $\operatorname{tg} \alpha: \operatorname{tg} \beta: \operatorname{tg} \gamma=1: 3: 12$, it follows that $\operatorname{tg} \alpha=k, \operatorname{tg} \beta=3 k, \operatorname{tg} \gamma=12 k$.

Substituting into the previous equality, we get:

$$
16 k=36 k^{3} \Longrightarrow k^{2}=\frac{4}{9} \Longrightarrow k=\frac{2}{3},
$$

so

$$
\operatorname{tg} \alpha=\frac{2}{3}, \quad \operatorname{tg} \beta=2, \quad \operatorname{tg} \gamma=8
$$

It follows that

$\sin \alpha=\frac{\operatorname{tg} \alpha}{\sqrt{1+\operatorname{tg}^{2} \alpha}}=\frac{2}{\sqrt{13}}, \quad \sin \beta=\frac{\operatorname{tg} \beta}{\sqrt{1+\operatorname{tg}^{2} \beta}}=\frac{2}{\sqrt{5}}, \quad \sin \gamma=\frac{\operatorname{tg} \gamma}{\sqrt{1+\operatorname{tg}^{2} \gamma}}=\frac{8}{\sqrt{65}}$,

thus

$$
\sin \alpha \cdot \sin \beta \cdot \sin \gamma=\frac{2}{\sqrt{13}} \cdot \frac{2}{\sqrt{5}} \cdot \frac{8}{\sqrt{65}}=\frac{32}{65}
$$