## Task B-3.4.

For the lengths of the legs of a right triangle $a$ and $b$, the following equality holds:

$$
\log (a+b)=\frac{1}{2} \cdot \log b+\frac{1}{2} \cdot \log (a+3 b)
$$

Calculate the measure of the angle opposite the leg of length $a$.
## Solution.

Using the rules for logarithms, we get

$$
\begin{array}{ll}
\log (a+b)=\frac{1}{2} \cdot \log [b \cdot(a+3 b)] & \\
\log (a+b)^{2}=\log [b \cdot(a+3 b)] & 1 \text { point } \\
(a+b)^{2}=b \cdot(a+3 b) . & 1 \text { point }
\end{array}
$$

From this, we have $a^{2}+a b-2 b^{2}=0$.

After dividing by $b^{2}$, we get

$$
\begin{array}{r}
\left(\frac{a}{b}\right)^{2}+\frac{a}{b}-2=0 \text { or } \operatorname{tg}^{2} \alpha+\operatorname{tg} \alpha-2=0 \\
\operatorname{tg} \alpha=1 \text { or } \operatorname{tg} \alpha=-2
\end{array}
$$

In a right-angled triangle, the solution can only be $\operatorname{tg} \alpha=1$, i.e., $\alpha=45^{\circ}$.

1 point