Question

a pilot is flying over the ocean. she determines that the angles of depression to two ships are 64° and 53°, as shown in the figure below. the plane is 5 miles from the ship located at point a. how far apart are the ships? round your answer to the nearest tenth of a mile.

Explanation:

Step1: Find the angle at the plane

Since the angles of depression are \(53^{\circ}\) and \(64^{\circ}\), the angle at the plane in the triangle is \(180^{\circ}-53^{\circ}-64^{\circ}=63^{\circ}\).

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}\). Let the distance between the ships be \(x\). We know one side \(a = 5\) (distance from plane to ship \(A\)), angle \(A = 64^{\circ}\), and angle \(B=63^{\circ}\). So \(\frac{x}{\sin64^{\circ}}=\frac{5}{\sin63^{\circ}}\).

$$x=\frac{5\times\sin64^{\circ}}{\sin63^{\circ}}$$

$$x=\frac{5\times0.8988}{0.8910}\approx5.1$$

Answer:

\(5.1\)