17. Given a triangle $\triangle A B C, \angle A B C=80^{\circ}, \angle A C B=70^{\circ}$ and $B C=2$. A perpendicular line is drawn from $A$ to $B C$, another perpendicular line drawn from $B$ to $A C$. The two perpendicular lines meet at $H$. Find the length of AH.
(2 marks)
17. 在 $\triangle A B C$ 中, $\angle A B C=80^{\circ}, \angle A C B=70^{\circ}$, 且 $B C=2$ 。由 $A$ 作線垂直於 $B C$, 由 $B$ 作線垂直於 $A C$,兩垂直線相交於 $H$ 。求 $A H$ 的長度。
17. $2 \sqrt{3}$
17. Note that $H$ is the orthocentre of $\triangle A B C$. We have $\angle H B C=20^{\circ}, \angle H C B=10^{\circ}, \angle B A H=80^{\circ}$ and $\angle A B H$ $=\angle A C H=60^{\circ}$. Applying sine law in $\triangle B H C$, we have
$$
\begin{aligned}
\frac{2}{\sin 150^{\circ}} & =\frac{B H}{\sin 10^{\circ}} \\
B H & =4 \sin 10^{\circ}
\end{aligned}
$$

Applying sine law in $\triangle A B H$, we have
$$
\begin{aligned}
\frac{4 \sin 10^{\circ}}{\sin 10^{\circ}} & =\frac{A H}{\sin 60^{\circ}} \\
A H & =2 \sqrt{3}
\end{aligned}
$$

Alternative Solution
Note that $H$ is the orthocentre of $\triangle A B C$. Construct the circumcircle of $\triangle A B C$ and the diameter $B D$. The diameter of the circle is
$$
\frac{B C}{\sin A}=\frac{2}{\sin 30^{\circ}}=4 .
$$

Since $A H$ and $D C$ are both perpendicular to $B C$, they are parallel. Similarly, $A D$ and $H C$ are parallel. So $A H C D$ is a parallelogram, and thus
$$
\begin{aligned}
A H & =D C=\sqrt{B D^{2}-B C^{2}} \\
& =\sqrt{4^{2}-2^{2}} \\
& =2 \sqrt{3}
\end{aligned}
$$