Question

two ships a and b leave port at the same time, ship a sailing on a course bearing 023 degrees at a speed of 11 km/h, and ship b sailing on a course bearing 113 degrees at 15 km/h. find the bearing to the nearest degree from ship b to ship a one hour later.

Explanation:

Step1: Calculate the distances

Since distance \(d = vt\) (where \(v\) is speed and \(t\) is time), after \(t = 1\) hour, \(d_A=11\times1 = 11\) km and \(d_B = 15\times1=15\) km.
The angle between their paths \(\theta=113 - 23=90^{\circ}\)

Step2: Use the Pythagorean theorem to find the distance between the ships

Let the distance between the ships be \(d\). By the Pythagorean theorem \(d=\sqrt{11^{2}+15^{2}}=\sqrt{121 + 225}=\sqrt{346}\approx18.6\) km

Step3: Use trigonometry to find the angle

Let \(\alpha\) be the angle we need to find. \(\tan\alpha=\frac{11}{15}\), so \(\alpha=\arctan(\frac{11}{15})\approx36.59^{\circ}\)

Step4: Calculate the bearing

The bearing from \(B\) to \(A\): The bearing of \(B\) is \(113^{\circ}\). The angle between the line from \(B\) to \(A\) and the path of \(B\) is \(\alpha\). The bearing \(=113^{\circ}+ 180^{\circ}-\alpha\approx113 + 180- 36.59=301^{\circ}\)

Answer:

\(301^{\circ}\)