Question #136182
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.
1
Expert's answer
2020-10-06T18:24:28-0400



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

AC=152=30AC=15\cdot 2=30 (nautical miles), AB=202=40AB=20\cdot 2=40 (n. m.).

ABC:BAC=12535=90.\triangle ABC: \angle BAC=125^{\circ}-35^{\circ}=90^{\circ}.


a)

The distance between them is BC=AB2+AC2=302+402=50BC=\sqrt{AB^2+AC^2}=\sqrt{30^2+40^2}=50 (n. m.).


c)

OBA=35,ABC=arcsin3050=arcsin3537.\angle OBA=35^{\circ}, \angle ABC=\arcsin{\frac{30}{50}}=\arcsin{\frac{3}{5}}\approx 37^{\circ}.

y=EBC=1803537=108.y= \angle EBC=180^{\circ}-35^{\circ}-37^{\circ}=108^{\circ}.

b)

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

x=360FCB=36072=288.x=360^{\circ}- \angle FCB=360^{\circ}- 72^{\circ}=288^{\circ}.

Answer: 50 nautical miles, 288,108.288^{\circ}, 108^{\circ}.


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