X = Vw = -85 km/h.
Y = Vp = 230 km/h.
a. Vr = sqrt(X^2 + Y^2) =
b. Tan A = Y/X = 230/-85 = -2.70588.
A =-69.7o=69.7o N. of W. = 20.3o W. of
N. = Bearing.
Y = Vp = 230 km/h.
a. Vr = sqrt(X^2 + Y^2) =
b. Tan A = Y/X = 230/-85 = -2.70588.
A =-69.7o=69.7o N. of W. = 20.3o W. of
N. = Bearing.
We can use vector addition to determine the resultant direction of the plane's motion. The direction of the plane is the angle formed between its motion due north and the wind vector due west.
Using the trigonometric concept of tangent, we can find the angle (θ) from the north:
θ = arctan(wind velocity/plane velocity)
θ = arctan(85/230)
Using a calculator, we find that θ ≈ 20.3 degrees.
Therefore, the plane is flying at approximately 20.3 degrees west of due north.
To calculate the actual speed of the plane, we need to find the resultant velocity. We can use vector addition again to determine this.
By using the Pythagorean theorem, we find:
Resultant velocity = √(plane velocity^2 + wind velocity^2)
= √(230^2 + 85^2)
= √(52900 + 7225)
= √(60125)
≈ 245 km
Hence, the actual speed of the plane is approximately 245 km/h.
Finally, the bearing from the north can be calculated by subtracting the angle (20.3 degrees) from 90 degrees (which represents due east):
Bearing from the north = 90 - 20.3
= 69.7 degrees
Therefore, the bearing from the north is approximately 69.7 degrees.