18. In $\triangle A B C, A B=13, B C=14$ and $C A=15 . P$ is a point inside $\triangle A B C$ such that $\angle P A B=\angle P B C=\angle P C A$. Find $\tan \angle P A B$.
(2 marks)
在 $\triangle A B C$ 中, $A B=13 、 B C=14 、 C A=15 \circ P$ 是 $\triangle A B C$ 內的一點, 使得 $\angle P A B$ $=\angle P B C=\angle P C A$ 。求 $\tan \angle P A B$ 。
18. $\frac{168}{295}$

Applying cosine law in $\triangle P A B, \triangle P B C$ and $\triangle P C A$ respectively, we have
$$
\begin{array}{l}
b^{2}=a^{2}+13^{2}-2(a)(13) \cos x \\
c^{2}=b^{2}+14^{2}-2(b)(14) \cos x \\
a^{2}=c^{2}+15^{2}-2(c)(15) \cos x
\end{array}
$$

Adding up the three equations and simplifying, we get
$$
2(13 a+14 b+15 c) \cos x=13^{2}+14^{2}+15^{2} \text {. }
$$

It follows that $\tan x=\frac{\sin x}{\cos x}=\frac{168}{13 a+14 b+15 c} \cdot \frac{2(13 a+14 b+15 c)}{13^{2}+14^{2}+15^{2}}=\frac{168}{295}$.
18. The semi-perimeter of $\triangle A B C$ is $(13+14+15) \div 2=21$ and hence $\triangle A B C$ has area
$$
\sqrt{21(21-13)(21-14)(21-15)}=84
$$
by Heron's formula. Let $\angle P A B=\angle P B C=\angle P C A=x$, $P A=a, P B=b$ and $P C=c$. Then we also have
$$
84=\frac{1}{2}(13)(a) \sin x+\frac{1}{2}(14)(b) \sin x+\frac{1}{2}(15)(c) \sin x
$$
and hence $(13 a+14 b+15 c) \sin x=168$.