\left.\begin{array}{ll}{[} & \text { The Law of Sines } \\ {[\text { Opposite the larger side lies the larger angle }]}\end{array}\right]

In triangle $ABC$, sides $AB$ and $AC$ are equal to $\sqrt{10}$ and $\sqrt{2}$, respectively, and the radius of the circumcircle of triangle $ABC$ is $\sqrt{5}$. Find the side $BC$ and the angle $ACB$, given that angle $ACB$ is acute.
Apply the formula $a=2 R \sin \boldsymbol{\alpha}$.

## Solution

Note that $\sin \angle A C B=\frac{A B}{2 R}$ (where $R=\sqrt{5}$ is the radius of the circumcircle of triangle $A B C$), i.e., $\sin \angle A C B=\frac{\sqrt{2}}{2}$. Therefore, $\angle A C B=45^{\circ}$. Similarly, we find that $\sin \angle A B C=\frac{1}{\sqrt{10}}$.

Since $A C<A B$ and angle $A C B$ is acute, angle $A B C$ is also acute. Therefore, $\cos \angle A B C=\frac{3}{\sqrt{10}}$. Then,

\[
\begin{gathered}
\sin \angle A=\sin \left(180^{\circ}-\angle B-\angle C\right)=\sin (\angle B+\angle C)= \\
=\sin \angle B \cos \angle C+\cos \angle B \sin \angle C=\frac{\sqrt{2}}{2} \cdot \frac{1}{\sqrt{10}}+\frac{\sqrt{2}}{2} \cdot \frac{3}{\sqrt{10}}=\frac{2}{\sqrt{5}}
\end{gathered}
\]

Therefore,

\[
B C=2 R \sin \angle A=2 \sqrt{5} \cdot \frac{2}{\sqrt{5}}=4
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-57.jpg?height=441&width=460&top_left_y=1689&top_left_x=799)

## Answer

$4 ; 45^{\circ}$.