## Task B-3.3.

In triangle $ABC$, $\frac{|AB|}{|AC|}=\frac{\sqrt{2}+\sqrt{6}}{2}$ and $\angle BAC=45^{\circ}$. Determine the measures of the remaining two angles of triangle $ABC$.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_c317f9bfd22736a3ecebg-23.jpg?height=399&width=563&top_left_y=346&top_left_x=678)

With the labels as in the figure, we have $\frac{c}{b}=\frac{\sqrt{2}+\sqrt{6}}{2}$ and $\alpha=45^{\circ}$.

Substituting $c=\frac{\sqrt{2}+\sqrt{6}}{2} b$ into the cosine rule, we get the following sequence of equalities:

$$
\begin{gathered}
a^{2}=b^{2}+c^{2}-2 b c \cos \alpha \\
a^{2}=b^{2}+\left(\frac{\sqrt{2}+\sqrt{6}}{2} b\right)^{2}-2 b \frac{\sqrt{2}+\sqrt{6}}{2} b \cos 45^{\circ} \\
a^{2}=b^{2}+\frac{8+2 \sqrt{12}}{4} b^{2}-\frac{2+\sqrt{12}}{2} b^{2} \\
a^{2}=b^{2}+\frac{4+\sqrt{12}}{2} b^{2}-\frac{2+\sqrt{12}}{2} b^{2} \\
a^{2}=b^{2}+b^{2} \\
a^{2}=2 b^{2}
\end{gathered}
$$

Thus, $a=\sqrt{2} b$.

Applying the sine rule, we get

$$
\sin \beta=\frac{b \sin \alpha}{a}=\frac{b \sin 45^{\circ}}{\sqrt{2} b}=\frac{1}{2}
$$

From $\sin \beta=\frac{1}{2}$ and $\alpha=45^{\circ}$, we conclude that the angle $\beta$ must be an acute angle, i.e., $\beta=30^{\circ}$.

1 point

$\gamma=180^{\circ}-\alpha-\beta=180^{\circ}-45^{\circ}-30^{\circ}=105^{\circ}$.

1 point