Answer to Question #135308 in Trigonometry for Galvez Yao

Question

Two ships leave a port at the same time. The first ship sails on a course of 35°at 15 knots while the second ship sails on a course of 125°at 20 knots. After 2 hours find the: (a) distance between the two ships; (b) bearing of the first ship from the second ship; and (c) bearing of the second ship from the first ship.

Answer 1

(a) the angle between courses is 125° - 35° = 90°.

After 2 hours the first ship moves $15\cdot 2 = 30$ hknots and the second moves $20\cdot 2 = 40$ hknots.

The distance can be calculated through the application of the Pythagorean theorem because the distances between the ships and between each ship and the port form rectangular triangle: $\sqrt{30^2+40^2} = 50$ hknots.

(b) The bearing of the first ship from the second ship is te angle between the direction to the north and to the first ship:

$360^\circ -\angle EDC = 360^\circ - (180^\circ - 90^\circ - \angle ECD), \\ \angle ECD = 180^\circ - \angle BCD - \angle ACB = 180^\circ -\arcsin\dfrac{BD}{CD} - (180^\circ - 90^\circ - 35^\circ) = 180^\circ - 53^\circ - 55^\circ = 72^\circ.\\ 360^\circ - (180^\circ - 90^\circ - \angle ECD) = 360^\circ - (180^\circ - 90^\circ -72^\circ) = 360^\circ - 18^\circ = 342^\circ.$

(c) The bearing of the second ship from the first ship is

$90^\circ + \angle ECD = 90^\circ + (180^\circ - \angle ACB - \angle BCD) = 90^\circ +(180^\circ -55^\circ - 53^\circ) = 162^\circ.$