8,9 In triangle $A B C$, angle $C$ is equal to $60^{\circ}$, and the bisector of angle $C$ is $5 \sqrt{3}$. The lengths of sides $A C$ and $B C$ are in the ratio 5:2, respectively. Find the tangent of angle $A$ and side $B C$.
Let $C D$ be the bisector of a given triangle. Denote $B C=2 t, A C=5 t$. Find $t$ from the equation

## Solution

Let $C D=5 \sqrt{3}$ be the bisector of the given triangle. Denote $B C=2 t, A C=5 t$. Then

$$
\begin{gathered}
S_{\triangle \mathrm{ABC}}=\frac{1}{2} \cdot B C \cdot A C \sin 60^{\circ}=\frac{5}{2} t^{2} \sqrt{3} \\
S_{\triangle \mathrm{ABC}}=S_{\triangle \mathrm{BCD}}+S_{\triangle \mathrm{ACD}}=\frac{1}{2} \cdot B C \cdot C D \sin 30^{\circ}+\frac{1}{2} \cdot A C \cdot C D \sin 30^{\circ}= \\
=\frac{1}{2} \cdot 2 t \cdot 5 \sqrt{3} \cdot \frac{1}{2}+\frac{1}{2} \cdot 5 t \cdot 5 \sqrt{3} \cdot \frac{1}{2}=\frac{1}{2} \cdot 7 t \cdot 5 \sqrt{3} \cdot \frac{1}{2}=\frac{35}{4} t \sqrt{3}
\end{gathered}
$$

From the equation $\frac{5}{2} t^{2} \sqrt{3}=\frac{35}{4} t \sqrt{3}$, we find that $t=\frac{7}{2}$. Then $B C=2 t=7, A C=5 t=\frac{35}{2}$.

Let $E$ be the projection of point $D$ onto the line $A C$. Since $B C<A C$, then $\angle D A C=\angle B A C<90^{\circ}$, so point $E$ lies on the side $A C$, not on its extension. Then from the right triangles $C E D$ and $A E D$, we find that

$$
\begin{gathered}
D E=C D \sin 30^{\circ}=\frac{1}{2} C D=\frac{5 \sqrt{3}}{2}, C E=C D \cos 30^{\circ}=5 \sqrt{3} \cdot \frac{\sqrt{3}}{2}=\frac{15}{2}, \\
A E=A C-C E=\frac{35}{2}-\frac{15}{2}=10, \operatorname{tg} \angle D A E=\frac{D E}{A E}=\frac{\frac{5 \sqrt{3}}{2}}{10}=\frac{\sqrt{3}}{4} .
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_8027e864227276d12944g-40.jpg?height=278&width=532&top_left_y=1313&top_left_x=769)

Answer

$\frac{\sqrt{3}}{4} ; 7$