14. (2 marks) In $\triangle A B C, \angle A C B=3 \angle B A C, B C=5, A B=11$. Find $A C$.
(2 分) 在 $\triangle A B C$ 中 $\angle A C B=3 \angle B A C, B C=5, A B=11$ 。求 $A C$ 。
14. From $C$ construct $\angle A C D=\angle B A C$, with $D$ on $A B$. Then $\angle B D C=$ $\angle B C D=2 \angle D A C$. Hence $B C=B D=5, A D=C D=6$. Apply cosine rule to $\triangle B D C$, get $\angle D B C=\frac{5^{2}+5^{2}-6^{2}}{2 \times 5 \times 5}=\frac{14}{50}$. Apply cosine rule to $\triangle A B C$, get $A C^{2}=11^{2}+5^{2}-2(5)(11) \cos \angle B=146-\frac{154}{5}=\frac{576}{5}$, Hence $A C=\frac{24}{\sqrt{5}}=\underline{\underline{\frac{24 \sqrt{5}}{5}}}$.

Alternate Solution

Let $\angle C A B=x<90^{\circ}$. By the sine rule, we have $\frac{11}{\sin 3 x}=\frac{5}{\sin x}=\frac{A C}{\sin \left(180^{\circ}-x\right)}$.
From the first equality, we have $\frac{11}{5}=\frac{\sin 3 x}{\sin x}=3-4 \sin ^{2} x$, hence $\sin x=\frac{1}{\sqrt{5}}$, and $\cos x=\frac{2}{\sqrt{5}}$.
From the second equality, we have $A C=\frac{5 \sin 4 x}{\sin x}=5\left(4 \cos x\left(1-2 \sin ^{2} x\right)\right)$
$$
=(5)(4)\left(\frac{2}{\sqrt{5}}\right)\left(1-\frac{2}{5}\right)=\frac{24}{\sqrt{5}}=\underline{\underline{\frac{24 \sqrt{5}}{5}}}
$$
(The compound angle formula $\sin (A+B)=\sin A \cos B+\cos A \sin B$ is used repeatedly here.)