4. In triangle $ABC$, $\overline{AB}=\overline{AC}$ and $\frac{\cos \angle BAC}{\cos \angle ABC}=\frac{7}{15}$. Calculate $\frac{\sin \angle BAC}{\sin \angle ABC}$.
Solution. Let $\overline{A B}=c, \overline{B C}=a, \overline{A C}=b, \angle B A C=\alpha, \angle A B C=\beta$ and $b=c$. From the cosine theorem, it follows that

$$
\cos \alpha=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=1-\frac{a^{2}}{2 b^{2}} \quad \text { and } \quad \cos \beta=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{a}{2 b} .
$$

From the condition $\frac{\cos \angle B A C}{\cos \angle A B C}=\frac{7}{15}$, it follows that

$$
\begin{aligned}
0 & =15 \cos \alpha-7 \cos \beta=15\left(1-\frac{a^{2}}{2 b^{2}}\right)-7 \frac{a}{2 b} \\
& =-\frac{15}{2}\left(\frac{a}{b}\right)^{2}-\frac{7}{2} \frac{a}{b}+15=\left(-\frac{5}{2} \frac{a}{b}+3\right)\left(3 \frac{a}{b}+5\right)
\end{aligned}
$$

Since $a, b>0$, it follows that $-\frac{5}{2} \frac{a}{b}+3=0$, or $\frac{a}{b}=\frac{6}{5}$. From the sine theorem, it follows that $\frac{\sin \alpha}{\sin \beta}=\frac{6}{5}$.