Two ships leave a port at 9 a.m. One travels at a bearing of N 53° W at 13 miles per hour, and the other travels at a bearing of S 67° W at s miles per hour.

Question

Two ships leave a port at 9 a.m. One travels at a bearing of N 53° W at 13 miles per hour, and the other travels at a bearing of S 67° W at s miles per hour.

Find the speed s that the second ship must travel so that the ships are 44 miles apart at noon. (Round your answer to two decimal places.) 

___mi/h

Answer 1

θ = 180° - (53° + 67°) = 60°; a = 13·3 = 39 mi; 44/sin60ˆ = 39/sinα; sinα = 39sin60°/44 = 0.7676;

α = 50°; β = 180° - (50° + 60°) = 70°; 44/sin60° = b/sinsin70°; b = 44sin70°/sin60° = 47.74 mi

Speed s = 47.74/3 = 15.9 mi/h