Answer to Question #136182 in Trigonometry for jay

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

Let ships start from A. After 2 hours first ship will be in С and second ship will be in B.

$AC=15\cdot 2=30$ (nautical miles), $AB=20\cdot 2=40$ (n. m.).

$\triangle ABC: \angle BAC=125^{\circ}-35^{\circ}=90^{\circ}.$

a)

The distance between them is $BC=\sqrt{AB^2+AC^2}=\sqrt{30^2+40^2}=50$ (n. m.).

c)

$\angle OBA=35^{\circ}, \angle ABC=\arcsin{\frac{30}{50}}=\arcsin{\frac{3}{5}}\approx 37^{\circ}.$

$y= \angle EBC=180^{\circ}-35^{\circ}-37^{\circ}=108^{\circ}.$

b)

$\angle FCB=180^{\circ}- \angle EBC=72^{\circ}.$

$x=360^{\circ}- \angle FCB=360^{\circ}- 72^{\circ}=288^{\circ}.$

Answer: 50 nautical miles, $288^{\circ}, 108^{\circ}.$