Question

two ships a and b leave port at the same time, ship a sailing on a course bearing 040 degrees at a speed of 12 km/h, and ship b sailing on a course bearing 130 degrees at 16 km/h. find the bearing from ship b to ship a one hour later.

Explanation:

Step1: Calculate the distances

Since distance \(d = vt\), for \(t = 1h\), \(d_A=12\times1 = 12km\), \(d_B = 16\times1=16km\). The angle between their paths \(\theta=130 - 40=90^{\circ}\).

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

Let the distance between \(A\) and \(B\) be \(c\). By \(c^{2}=a^{2}+b^{2}\) (\(a = 12\), \(b = 16\)), \(c=\sqrt{12^{2}+16^{2}}=\sqrt{144 + 256}=\sqrt{400}=20km\).

Step3: Use trigonometry to find the angle

Let \(\alpha\) be the angle. \(\tan\alpha=\frac{12}{16}=0.75\), \(\alpha=\arctan(0.75) = 36.87^{\circ}\).

Step4: Calculate the bearing

The bearing of \(A\) from \(B\): The direction of \(B\) is \(130^{\circ}\). The bearing \(=130^{\circ}+ 180^{\circ}- 36.87^{\circ}=301.13^{\circ}\approx301^{\circ}\)

Answer:

\(301^{\circ}\)