2. Zadatak

Kako su kutovi trokuta šiljasti, to su njihovi tangensi istog predznaka i vrijedi

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

Slijedi

$$\operatorname{tg}\gamma-\operatorname{tg}\gamma\cdot\operatorname{tg}\alpha\cdot\operatorname{tg}\beta=-\operatorname{tg}\alpha-\operatorname{tg}\beta,$$

odnosno

$$\operatorname{tg}\gamma+\operatorname{tg}\alpha+\operatorname{tg}\beta=\operatorname{tg}\alpha\cdot\operatorname{tg}\beta\cdot\operatorname{tg}\gamma.$$

Kako je $\operatorname{tg}\alpha:\operatorname{tg}\beta:\operatorname{tg}\gamma=1:3:12$, slijedi $\operatorname{tg}\alpha=k$, $\operatorname{tg}\beta=3k$, $\operatorname{tg}\gamma=12k$. Tada uvrštavanjem u prethodnu jednakost dobivamo

$$16k=36k^3\Longrightarrow k^2=\frac49\Longrightarrow k=\frac23,$$

te je

$$\operatorname{tg}\alpha=\frac23,\qquad \operatorname{tg}\beta=2,\qquad \operatorname{tg}\gamma=8.$$

Slijedi

$$\sin\alpha=\frac{\operatorname{tg}\alpha}{\sqrt{1+\operatorname{tg}^2\alpha}}=\frac2{\sqrt{13}},\quad \sin\beta=\frac{\operatorname{tg}\beta}{\sqrt{1+\operatorname{tg}^2\beta}}=\frac2{\sqrt5},\quad \sin\gamma=\frac{\operatorname{tg}\gamma}{\sqrt{1+\operatorname{tg}^2\gamma}}=\frac8{\sqrt{65}},$$

pa je

$$\sin\alpha\cdot\sin\beta\cdot\sin\gamma=\frac2{\sqrt{13}}\cdot\frac2{\sqrt5}\cdot\frac8{\sqrt{65}}=\frac{32}{65}.$$