Task B-2.2. In a right-angled triangle with acute angles $\alpha$ and $\beta$, it is given that $\operatorname{tg} \alpha=2$. Calculate the exact value of the expression

$$
\frac{3 \sin \alpha+2 \sin \beta-\sqrt{5}}{3 \cos \alpha-2 \cos \beta+\sqrt{5}}
$$
First solution. All triangles with the given property are similar to the drawn triangle $A B C$.

![](https://cdn.mathpix.com/cropped/2024_05_30_7c0a9615fb5cd46ea1c5g-06.jpg?height=499&width=325&top_left_y=1995&top_left_x=865)

If $\operatorname{tg} \alpha=2$, that is, if the ratio of the legs is $2: 1$, then the leg $a=2 k$, the leg $b=k$, and the hypotenuse $c=\sqrt{5} k$.

(2 points)

Note: If these facts are only marked on the figure without explanation, two points are still given. The parameter $k$ can be omitted.

The given expression is then equal to

$$
\frac{3 \sin \alpha+2 \sin \beta-\sqrt{5}}{3 \cos \alpha-2 \cos \beta+\sqrt{5}}=\frac{3 \frac{2}{\sqrt{5}}+2 \frac{1}{\sqrt{5}}-\sqrt{5}}{3 \frac{1}{\sqrt{5}}-2 \frac{2}{\sqrt{5}}+\sqrt{5}}=\frac{\frac{8}{\sqrt{5}}-\sqrt{5}}{-\frac{1}{\sqrt{5}}+\sqrt{5}}=\frac{3}{4}
$$