A ship leaves port P on a bearing of 125(true bearing) for 25 km. It then changes course to S10 W for 10 km a) Find the distance and the bearing of the ship from Port P

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justaguide | College Teacher | (Level 2) Distinguished Educator

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A ship leaves port P on a bearing of 125 (true bearing) for 25 km. It then changes course to S10 W for 10 km.

The distance traveled by the ship towards the East is 25*cos 35 - 10*sin 10 = 18.74. The distance traveled by the ship towards the South is 25*sin 35 + 10*cos 10 = 24.187

The final distance of the ship from the port P is `sqrt(18.74^2 + 24.18^2) ~~ 30.59` km

The bearing of the ship `90 + tan^-1( 24.18/18.74)` = 142

At the final point the ship is 30.59 km from port P and the true bearing of the ship is 142

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