Aship moves through the water at 30 miles/hour at an angle of 30° south of east. the water is moving 5 miles/hour at an angle of 20° east of north. identify the ship's vector, the water current's vector, and the vector representing the ship's actual motion.

If you want (x, y) coordinates to match (East, North) coordinates, then your vectors are
.. ship = (30 mph, ∠-30°)
.. water = (5 mph, ∠70°)
.. ship +water = (29.545, ∠-20.406°) . . . . . from my graphing calculator
That is, the total is 29.5 mph E20°S.

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If you are required to do the calculation by hand, you add the components of each vector. They are computed from
.. (r, θ) = r*(cos(θ), sin(θ))
For example, the ship's speed would be
.. (30 mph, ∠-30°) = (30 mph)*(cos(-30°), sin(-30°)) = (25.9808, -15.0000) mph
and the total of ship and water would be
.. (27.6909, -10.3015) mph
This is converted back to (r, θ) by
.. r = √(x^2 +y^2)
.. θ = arctan(y/x) . . . . with attention to quadrant

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The above calculations use (E, N) coordinates so north is in the direction of +y, and angles are measured counterclockwise from +x.

Navigation problems can also be worked using (N, E) coordinates with angles measured the way bearings are conventionally measured, clockwise from north. The math is exactly the same, and the conversion to and from rectangular (N, E) coordinates to polar (r, bearing) coordinates is virtually identical (bearing = arctan(E/N)). I find it less work to do it this way and not have to worry about numbers getting swapped in translation.

For this problem, I would use ship=(30,∠120°), current=(5,∠20°), total=(29.5∠110.4°), where all the angles are bearing angles measured CW from north.

Igot 108 hope that’s right !