24 A helicopter $H$ is hovering above a straight, horizontal road $A B$ of length $600 \mathrm{~m}$. The angles of elevation of $H$ from $A$ and $B$ are $7^{\circ}$ and $13^{\circ}$ respectively. The point $C$ lies on the road directly below $H$. a Use the sine rule to show that $H B=\frac{600 \sin 7^{\circ}}{\sin 160^{\circ}}$. b Hence find the height $\mathrm{CH}$ of the helicopter above the road, correct to the nearest metre.

Step 1 ：\(\triangle AHB\) and \(\triangle CHB\) are right triangles with \(\angle AHB = 7^\circ\) and \(\angle CHB = 90^\circ\)

Step 2 ：In \(\triangle AHB\), apply the sine rule: \(\frac{HB}{\sin{AHB}} = \frac{AB}{\sin{160^\circ}}\)

Step 3 ：Substitute the given values: \(\frac{HB}{\sin{7^\circ}} = \frac{600}{\sin{160^\circ}}\)

Step 4 ：Solve for \(HB\): \(HB = \frac{600 \sin{7^\circ}}{\sin{160^\circ}}\)

Step 5 ：In \(\triangle CHB\), use the tangent function: \(\tan{13^\circ} = \frac{CH}{HB}\)

Step 6 ：Substitute the expression for \(HB\): \(\tan{13^\circ} = \frac{CH}{\frac{600 \sin{7^\circ}}{\sin{160^\circ}}}\)

Step 7 ：Solve for \(CH\): \(CH = \frac{600 \sin{7^\circ}}{\sin{160^\circ}} \cdot \tan{13^\circ}\)

Step 8 ：Calculate the value of \(CH\) and round to the nearest meter: \(CH \approx 53\)

Step 9 ：\(\boxed{CH = 53}\)