SOLUTION: The heading of a plane is 27.7 degrees NE, and its air speed is 255 mi.h. If the wind is blowing from the south with a velocity of 42.0 mi/h, find the actual direction of travel o

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Question 524976: The heading of a plane is 27.7 degrees NE, and its air speed is 255 mi.h. If the wind is blowing from the south with a velocity of 42.0 mi/h, find the actual direction of travel of the plane, and its ground speed.
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
The heading of a plane is 27.7 degrees NE, and its air speed is 255 mi.h. If the wind is blowing from the south with a velocity of 42.0 mi/h, find the actual direction of travel of the plane, and its ground speed.
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ground speed = 292.84 mi/hr
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GS/sin(152.7) = 42/sin(A) A = the angle betwee the heading and the ground track.
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sin(A) = 42*sin(152.7)/GS
A = 3.77 degs
ground track = 27.7 - 3.77 = 23.93 degs, a heading of ~024
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Never seen fractions of degrees used in aviation.
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Update:
Edwin got 3.82 degs, only slightly different from 3.77, not significant.
But, he used the plane's heading as the x-axis, and so should have subtracted the 3.8... from the 27.7 heading to the the direction of the ground track.

Answer by Edwin McCravy(20055) (Show Source):
You can put this solution on YOUR website!
The heading of a plane is 27.7 degrees NE, and its air speed is 255 mi.h. If the wind is blowing from the south with a velocity of 42.0 mi/h, find the actual direction of travel of the plane, and its ground speed.

The other tutor used the law of cosines and the parallelogram method. I will use the component method: The red vector represents the wind's velocity. Since the wind is FROM the SOUTH, it is blowing TOWARD the NORTH! The green vector represents the plane's velocity. Note that that the angle 27.7� is measured clockwise from the vertical (North), so the angle we will use is measured counterclockwise from the horizontal (East) calculated as 90�-27.7� or 62.3�. x-components y-components Plane 255cos(90�) 255sin(90�) Wind 42cos(62.3�) 42sin(62.3�) ---------------------------------------------- Sums: 19.52336592 292.1865323 |Resultant| = = 292.8380635 mi/h Angle (counterclockwise from East) = tan^-1( = 86.177� To find the bearing clockwise from the North, we subtract from 90� and get 3.823�, and the bearing is written as N 3.823� E. I agree with the other tutor on the magnitude of the resultant, but we differ just a bit on the angle. Edwin