I don't know how to solve this.

Question

10. A ship is traveling due northeast at 18 knots with a current of 8 knots in the direction due north. What is the true course of the ship and what is its actual speed? (hint: course is measured from north clockwise)

Answer 1

Let W = <0, 8>  ... wind vector's x & y components

Let S = <12.7279, 12.7279>   ... Ship vector's x & y components based on isosceles right triangle, hypotenuse = 18 so each leg = 18 /√(2) = about 12.7279.

 

The result vector, formed by addition of components of the two individual vectors that are being added, is therefore

R = <12.7279, 20.7279.  Its magnitude is the square root of the sum of the squares of its two components square root of 591.646 = 24.32, approximately...that's the speed.

 

Now draw this resultant R vector in its standard position, in first quadrant.  You know all three sides, and so you can find angle θ = arctan (x/y) = arctan (12.7279/20.7279) = arctan .61405 =31.55 degrees.  Direction: 33.55 deg east of North, speed 24.32 knots.