Question

Two ships leave a port at the same time. The first ship sails on a course of 35 degrees at 15 knots while the second ship sails on a course of 125 degrees at 20 knots. Find after 2 hours (a) the distance between the ships, (b) the bearing from the first ship to the second ship, and (c) the bearing of the second ship to the first.

Answer 1

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Answer on Question #44125 – Math – Trigonometry

Two ships leave a port at the same time. The first ship sails on a course of 35 degrees at 15 knots while the second ship sails on a course of 125 degrees at 20 knots. Find after 2 hours (a) the distance between the ships, (b) the bearing from the first ship to the second ship, and (c) the bearing of the second ship to the first.

Solution:

#1

From the given headings we can say the angle between the courses of the two ships:

$$\alpha = 125{}^\circ - 35{}^\circ = 90{}^\circ$$

After $t = 2$ hours:

the 1st ship will have travelled distance

$$d_1 = v_1 \cdot t = 15 \text{ knots} \cdot 2\text{hours} = 30 \text{ hknots}$$

the 2nd ship will have travelled distance

$$d_2 = v_2 \cdot t = 20 \text{ knots} \cdot 2\text{hours} = 40 \text{ hknots}$$

We can use the law of cosines for a side/angle/side problem

$$a^2 = b^2 + c^2 - 2bc \cos \alpha$$

Assign the values as follows

a = distance between ships after 2 hours (side opposite angle $\alpha$)

b = 30 hknots

c = 40 hknots

$$\alpha = 90{}^\circ$$

Thus,

$$a^2 = 30^2 + 40^2 - 2 \cdot 30 \cdot 40 \cdot \cos 90{}^\circ$$ $$a = \sqrt{30^2 + 40^2} = 50 \text{ hknots}$$

#2 and #3

Now we can get the bearings or the other angles of the triangle by using (arc)trig functions.

Bearing from the first ship to the second ship:

$$\beta = \arctan \left(\frac{c}{b}\right) = \arctan \left(\frac{40}{30}\right) = 53{}^\circ$$

The bearing of the second ship to the first.

$$\gamma = 90{}^\circ - \beta = 37{}^\circ$$

Answer:

a) distance between two ships after 2 hours: 50 hknots

b) $53{}^\circ$

c) $37{}^\circ$