a plane is flying with airspeed of 140 mph and heading 280 degrees. The wind currents are running at 20mph at 260 degrees clockwise from due north. Find the true course and ground speed of the plane.

Vp=140mi/h @ 280 deg.,CCW. = Velocity of the plane.

Vw = 20mi/h @ 190 deg.CCW. = Velocity of the wind.

X = hor. = 140cos280 + 20cos190,
X = 24.2 + (-19.7) = 4.5mi/h.

Y = ver. = 140sin280 + 20sin190,
Y = -137.87 + (-3.47) = -141.34mi/h.

tanA = Y/X = -141.34/4.5 = - 31.41,
A = -88.2 deg,CW = 360 - 88.2 = 271.8
deg, CCW. = Direction.

R = X/cosA = 4.5/cos271.8 = 141.4mi/h = Magnatude.

To find the true course and ground speed of the plane, we need to take into account the effect of the wind on the plane.

1. Start by drawing a diagram to represent the situation. Draw a line representing the direction of the airplane's heading (280 degrees) and another line representing the direction of the wind (260 degrees clockwise from due north). Label the airspeed of the plane as 140 mph and the speed of the wind as 20 mph.

2. Calculate the wind velocity as the difference between the wind's direction and the direction opposite to the aircraft's heading. In this case, the difference is (280 - 180) = 100 degrees.

3. Use trigonometry to find the components of the wind velocity in the north-south and east-west directions. Since the wind direction is clockwise from due north, we need to use the sine and cosine functions accordingly. The north-south component is 20 mph * sin(100 degrees) ≈ -1.34 mph and the east-west component is 20 mph * cos(100 degrees) ≈ 3.23 mph.

4. Add the components of the wind velocity to the respective components of the airspeed of the plane to get the true velocity of the plane. In the north-south direction, the true velocity is 140 mph + (-1.34 mph) = 138.66 mph (southward). In the east-west direction, the true velocity is 0 mph + 3.23 mph = 3.23 mph (eastward).

5. Use the Pythagorean theorem to find the magnitude of the true velocity vector or the ground speed of the plane. The magnitude of the true velocity vector is √((138.66 mph)^2 + (3.23 mph)^2) ≈ 138.71 mph.

6. Finally, determine the angle of the true velocity vector or the true course of the plane using inverse trigonometric functions. The true course of the plane is arctan(3.23 mph / 138.66 mph) ≈ 1.32 degrees east of south.

Therefore, the true course of the plane is approximately 1.32 degrees east of south, and the ground speed of the plane is approximately 138.71 mph.