Problem 13. In triangle $A B C$, side $A C=$ $=5 \text{ cm}, B C-A B=2 \text{ cm}, \angle A: \angle C=2$. Find the lengths of sides $A B$ and $B C$.
Solution. Let $\angle A C B=\varphi$, then $\angle A=$ $=2 \varphi, \angle B=180^{\circ}-3 \varphi$. By the Law of Sines:

$$
\begin{gathered}
\frac{\sin 3 \varphi}{5}=\frac{\sin 2 \varphi}{B C}=\frac{\sin \varphi}{A B} \\
A B=\frac{5 \sin \varphi}{\sin 3 \varphi}, B C=\frac{5 \sin 2 \varphi}{\sin 3 \varphi}
\end{gathered}
$$

According to the problem statement

$$
\frac{5 \sin 2 \varphi}{\sin 3 \varphi}-\frac{5 \sin \varphi}{\sin 3 \varphi}=2
$$

Since $\sin 3 \varphi \neq 0$, equation (1) is equivalent to the equation

$$
\sin 2 \varphi-\sin \varphi=0.4 \sin 3 \varphi
$$

Using the formulas $\sin 2 \varphi=2 \sin \varphi \cos \varphi$ and $\sin 3 \varphi=$ $=3 \sin \varphi-4 \sin ^{3} \varphi$, we can rewrite equation (2) as: $2 \sin \varphi \cos \varphi-\sin \varphi=0.4\left(3 \sin \varphi-4 \sin ^{3} \varphi\right)$. Simplify the obtained equation:

$$
\begin{gathered}
2 \sin \varphi \cos \varphi-2.2 \sin \varphi+1.6 \sin ^{3} \varphi=0 \\
\sin \varphi\left(\cos \varphi-1.1+0.8 \sin ^{2} \varphi\right)=0 \\
\sin \varphi\left(\cos \varphi-1.1+0.8-0.8 \cos ^{2} \varphi\right)=0 \\
\quad \sin \varphi\left(0.8 \cos ^{2} \varphi-\cos \varphi+0.3\right)=0
\end{gathered}
$$

Equation (3) is equivalent to the equation

$$
0.8 \cos ^{2} \varphi-\cos \varphi+0.3=0
$$

Solving it, we get:

$$
\cos \varphi_{1}=0.75 ; \cos \varphi_{2}=0.5
$$

If $\cos \varphi_{2}=0.5$, then $\varphi_{2}=60^{\circ}, 2 \varphi_{2}=120^{\circ}$ and $\varphi_{2}+$ $+2 \varphi_{2}=180^{\circ}$, which means that $\varphi$ cannot be equal to $60^{\circ}$.

If $\cos \varphi_{1}=0.75$, then $\sin \varphi_{1}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$ and $\sin 2 \varphi=\frac{3 \sqrt{7}}{8} ; \sin 3 \varphi=3 \cdot \frac{\sqrt{7}}{4}-4 \cdot \frac{7 \sqrt{7}}{64}=\frac{5 \sqrt{7}}{16}$. Then

$$
A B=\frac{5 \cdot \frac{\sqrt{7}}{4}}{\frac{5 \sqrt{7}}{16}}=4 \quad(\mathrm{~cm}), B C=5 \cdot \frac{\frac{3 \sqrt{7}}{8}}{\frac{5 \sqrt{7}}{16}}=6 \quad(\mathrm{~cm})
$$