Question

A cruise ship maintains a speed of 13 knots (nautical miles per hour) sailing from San Juan to Barbados, a distance of 500 nautical miles. To avoid a tropical storm, the captain heads out of San Juan at a direction of 12° off a direct heading to Barbados. The captain maintains the 13-knot speed for 13 hours after which time the path to Barbados becomes clear of storms. (a) Through what angle should the captain turn to head directly to Barbados? b) Once the turn is made, how long will it be before the ship reaches Barbados if the same 13. knot speed is maintained? (a) The captain should head through an angle of (Do not round unt the final answer. Then round to one decimal place as needed.) (6) The time will take for the ship to reach Barbados is hours (Do not found until the final answer. Then found to one decimal place as needed.) Law ERT tab

Answer 1

Solution:

nr 600 Barbados 20° 7B ra The given figure is redesigned as the above figure, wher LADB = 90°, AB=600 Speed of ship = 13 knots, Time and 13 hours. ... Distance travelled = (speed & time) = AC In => AC = 13 x 13 - 169 nautical miles ABD sin 20 = BD BD = sin 20x AB - AB sÝ BD > sin 20 x 600 - 205.21 we know, AC= 169, To find CD we consider & ABD cos 20= AD = AD AB 600 2) AD= Cos20x600 2 563.81 As are to figure AD = ACACO A CD = AD - AC = 563.81 - 164 CD: 39408] To find the angle in which the captain turns for Borbades, we need to @ o find LBCD. Ching

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