Two ships, A and B, leave port simultaneously. Ship A travels northwest at 21 knots, while Ship B proceeds at 25 knots in a direction 32° west of south. Using 1 knot as equal to 1 nautical mile per hour: (a) We must calculate the magnitude of the velocity of Ship A relative to Ship B in knots. (b) Determine the direction of Ship A's velocity relative to Ship B, measured with respect to east. (c) We need to find out after how many hours the ships will be 180 nautical miles apart. (d) Finally, we'll determine the bearing of Ship B relative to Ship A at that time, using east as the positive x-direction and measuring north of east as a positive angle, with angles ranging from -180 degrees to 180 degrees, rounded to the nearest degree.

(a) To calculate the magnitude of the velocity of Ship A relative to Ship B, we should consider both velocities as vectors and then find the relative velocity vector. This is given by the vector subtraction of Ship B’s velocity vector from Ship A’s velocity vector.

For Ship A, traveling northwest is the same as traveling 45° west of north, since northwest is directly between north and west. Since we're considering east as the positive x-direction and north as the positive y-direction, we'll need to find components of Ship A's velocity vector along the south and the east direction (negative y and negative x components respectively).

For a speed of 21 knots at an angle of 45° west of north: V_Ax = -21 * cos(45°) (towards west, hence negative) V_Ay = 21 * sin(45°) (towards north, positive)

For Ship B, traveling 32° west of south will involve similar trigonometric considerations:

For a speed of 25 knots at an angle of 32° west of south: V_Bx = -25 * cos(32°) (towards west, hence negative) V_By = -25 * sin(32°) (towards south, negative)

The relative velocity (V_AB) of Ship A with respect to Ship B is calculated by subtracting the velocity vector of Ship B from the velocity vector of Ship A:

V_ABx = V_Ax - V_Bx V_ABy = V_Ay - V_By

Plugging in the values we have: V_ABx = -21 * cos(45°) - (-25 * cos(32°)) V_ABy = 21 * sin(45°) - (-25 * sin(32°))

The magnitude of the relative velocity (V_AB) is given by the Pythagorean theorem: V_AB = sqrt(V_ABx^2 + V_ABy^2)

Plugging in numbers, you would calculate that as follows.

(b) To find the direction of Ship A's velocity relative to Ship B, you need to find the angle that the velocity vector V_AB makes with respect to the positive x-direction (east). You can use the arctangent function to find this angle θ:

θ = arctan(V_ABy / V_ABx)

Please calculate the above values for the exact direction.

(c) To find out after how many hours the ships will be 180 nautical miles apart, we use the relative velocity magnitude and set up the equation:

Distance = Speed × Time

Here the distance is 180 nautical miles and the speed is |V_AB| which we calculated in part (a). Let's denote the time by t. So:

180 = |V_AB| * t

Solve for t by dividing both sides by |V_AB|:

t = 180 / |V_AB|

Please calculate the actual time using the magnitude from part (a).

(d) To find the bearing of Ship B relative to Ship A at the time they are 180 nautical miles apart, we will consider the positions they will have reached relative to each other by that time. Then we can calculate the bearing.

Initially, we found that the relative velocity vector angle θ was measured with respect to east. The bearing, however, needs to be north of east, so we must convert the angle we have to be measured from the north going clockwise.

If θ is positive, then the bearing from Ship A to Ship B would be 90° - θ (since it's measured from east and we need the angle from north). If θ is negative, then the bearing from Ship A to Ship B would be -(90° + θ) or 90° - |θ|.

You will then convert the angle into the bearing format by adjusting it within the range of -180 to 180 degrees, as asked.

Remember to calculate the exact values using the angles from part (b) and the actual relative velocity magnitude.