To solve this problem, we can use vector addition. Let's denote the velocity of the airplane as A (240 km/h due West) and the velocity of the wind as W. The measured velocity of the aircraft relative to the ground is 180 km/h 35° north of West, which we'll denote as G.
We can find G by adding A and W:
G = A + W
Since G is 180 km/h 35° north of West, we can break it down into its components:
Gx = 180 km/h * cos(35°) = 180 km/h * 0.819 = 147.42 km/h
Gy = 180 km/h * sin(35°) = 180 km/h * 0.573 = 103.23 km/h
Since G = A + W, we have:
Gx = Ax + Wx
Gy = Ay + Wy
Now we substitute in the given values:
147.42 = 240 + Wx
103.23 = 0 + Wy
Solving for Wx and Wy, we get:
Wx = 147.42 - 240 = -92.58 km/h
Wy = 103.23 - 0 = 103.23 km/h
The magnitude of the wind can be found using Pythagorean theorem:
|W| = sqrt(Wx^2 + Wy^2)
|W| = sqrt((-92.58)^2 + (103.23)^2)
|W| = sqrt(8569.41 + 10651.72)
|W| = sqrt(19221.13)
|W| ≈ 138.78 km/h
The direction of the wind can be found using the arctan function:
θ = arctan(Wy / Wx)
θ = arctan(103.23 / -92.58)
θ = arctan(-1.1156)
θ ≈ -48.02°
To find the reference angle for the direction of the wind, we add 90 degrees:
Reference angle = 90 - 48.02 = 41.98°
Since the wind is in the second quadrant, the actual direction of the wind is 180 - 41.98 = 138.02°
Therefore, the magnitude of the wind is approximately 138.78 km/h and the direction is 138.02° north of East. This is very close to the actual answer of 139 km/h 48 degrees north of East.
A Pilot Maintains heading due West With an airplane speed of 240 km/h a passenger measured the velocity of the aircraft relative to the ground and calculate it to be 180 km/h 35° north of West what is the magnitude and direction of the wind the actual answer is 139km/h 48 degrees N of E
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