Answer:
86.51° North of West or 273.49°
Explanation:
Let V' = velocity of boat relative to the earth, v = velocity of boat relative to water and V = velocity of water.
Now, by vector addition V' = V + v'.
Since v' = 6.10 m/s in the north direction, v' = (6.10 m/s)j and V = 100 m/s in the east direction, V = (100 m/s)i. So that
V' = V + v'
V' = (100 m/s)i + (6.10 m/s)j
So, we find the direction,Ф the boat must steer to from the components of V'.
So tanФ = 6.10 m/s ÷ 100 m/s
tanФ = 0.061
Ф = tan⁻¹(0.061) = 3.49°
So, the angle from the north is thus 90° - 3.49° = 86.51° North of West or 270° + 3.49° = 273.49°