Question

a ship sailing on a course bearing 015 degrees is 3500 m due south of a lighthouse. if the ship continues on this course, what is the closest distance the ship will come to the lighthouse? (round to the nearest whole number)

Explanation:

Step1: Analyze the angle

The bearing is \(015^{\circ}\), so the angle between the ship's path and the south - north line is \(15^{\circ}\).

Step2: Use the sine function

We know the initial distance of the ship from the lighthouse along the south - north line \(d = 3500\) m. The closest distance \(h\) of the ship to the lighthouse (perpendicular distance) can be found using the formula \(h=d\sin\theta\), where \(\theta = 15^{\circ}\) and \(d = 3500\) m.

$$h = 3500\times\sin(15^{\circ})$$

We know that \(\sin(15^{\circ})=\sin(45^{\circ}- 30^{\circ})=\sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.2588\)

$$h = 3500\times0.2588 = 905.8\approx906$$

Answer:

\(906\) m