## Task A-3.6.

Let $a, b$ and $c$ be the lengths of the sides of a triangle opposite to angles of size $\alpha, \beta$ and $\gamma$.

If $9 a^{2}+9 b^{2}=19 c^{2}$, determine $\frac{\operatorname{ctg} \gamma}{\operatorname{ctg} \alpha+\operatorname{ctg} \beta}$.
## First Solution.

We have

$$
\begin{array}{rlrl}
\operatorname{ctg} \alpha+\operatorname{ctg} \beta & =\frac{\cos \alpha}{\sin \alpha}+\frac{\cos \beta}{\sin \beta} & \\
& =\frac{\cos \alpha \sin \beta+\cos \beta \sin \alpha}{\sin \alpha \sin \beta}=\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta} & & 2 \text { points } \\
& =\frac{\sin \gamma}{\sin \alpha \sin \beta} . & 1 \text { point }
\end{array}
$$

It follows that

$$
\frac{\operatorname{ctg} \gamma}{\operatorname{ctg} \alpha+\operatorname{ctg} \beta}=\frac{\frac{\cos \gamma}{\sin \gamma}}{\frac{\sin \gamma}{\sin \alpha \sin \beta}}=\frac{\sin \alpha \sin \beta \cos \gamma}{\sin ^{2} \gamma}
$$

From the cosine rule, we have

$$
\cos \gamma=\frac{a^{2}+b^{2}-c^{2}}{2 a b}
$$

and from the sine rule, we have

$$
\frac{\sin \alpha}{\sin \gamma}=\frac{a}{c} \quad \text{and} \quad \frac{\sin \beta}{\sin \gamma}=\frac{b}{c}
$$

Substituting these, we get

$$
\frac{\operatorname{ctg} \gamma}{\operatorname{ctg} \alpha+\operatorname{ctg} \beta}=\frac{\sin \alpha \sin \beta \cos \gamma}{\sin ^{2} \gamma}=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \cdot \frac{a b}{c^{2}}=\frac{a^{2}+b^{2}-c^{2}}{2 c^{2}}
$$

Using the condition $a^{2}+b^{2}=\frac{19}{9} c^{2}$ from the problem, we get

$$
\frac{\operatorname{ctg} \gamma}{\operatorname{ctg} \alpha+\operatorname{ctg} \beta}=\frac{\frac{19}{9} c^{2}-c^{2}}{2 c^{2}}=\frac{5}{9}
$$