## Task 28/76

In a triangle with angles $\alpha, \beta, \gamma$ and the sides $a, b, c$ opposite to them, the following holds:

$$
\tan \alpha \cdot \tan \beta=3 \quad ; \quad \tan \beta \cdot \tan \gamma=6
$$

Calculate the angles $\alpha, \beta, \gamma$ and the side ratios $\frac{a}{c}$ and $\frac{b}{c}$.
It is

$$
\tan \gamma=\tan \left(180^{\circ}-\alpha-\beta\right)=-\tan (\alpha+\beta)=-\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}
$$

Thus, the given equations take the form

$$
\begin{array}{r}
\tan \alpha \tan \beta=3 \\
\tan \beta \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=-6
\end{array}
$$

Solving this system of equations, we obtain

$$
\tan \alpha=1 ; \quad \tan \beta=3 ; \quad \tan \gamma=2
$$

(negative tan-values are discarded, as $0^{\circ}<\alpha, \beta, \gamma<180^{\circ}$) which results in

$$
\begin{aligned}
& \frac{a}{b}=\sqrt{\frac{\tan ^{2} \alpha\left(1+\tan ^{2} \beta\right)}{\tan ^{2} \beta\left(1+\tan ^{2} \alpha\right)}}=\sqrt{\frac{1 \cdot(1+9)}{9 \cdot(1+1)}}=\sqrt{\frac{10}{18}} \approx 0.745 \\
& \frac{a}{c}=\sqrt{\frac{\tan ^{2} \alpha\left(1+\tan ^{2} \gamma\right)}{\tan ^{2} \gamma\left(1+\tan ^{2} \alpha\right)}}=\sqrt{\frac{1 \cdot(1+4)}{4 \cdot(1+1)}}=\sqrt{\frac{5}{18}} \approx 0.791
\end{aligned}
$$