Solving a Homework Assignment: Airplane Flight in a Windy Day

AI Thread Summary
An airplane with an airspeed of 580 km/h needs to maintain a heading of 38 degrees North of East while facing a 72 km/h wind from the north. To find the correct heading, vector components of both the airplane and wind must be considered. The initial calculations using trigonometric functions did not yield the expected angle of 43.6 degrees North of East. A suggestion was made to introduce an unknown for the heading and set up equations based on the law of sines or vector components. The discussion emphasizes the need to correctly account for the wind's effect on the airplane's trajectory to solve the problem accurately.
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Homework Statement


an airplane, whose air speed is 580km/h, is supposed to fly in a straight path 38N of E. But a steady 72km/h wind is blowing from the north. In what direction should the plane head? i know the answer is 43.6 N of E but i am not completely sure how to get it


Homework Equations


so i think i have to use 580cos(38) and 580sin(38) but when i do it this way it does not work.
My friend told me to use sin/72=sin(38+90)/580 which does work but i am not completely sure if that's what i should use

The Attempt at a Solution


before i used 580cos(38) 580sin(38) 72cos(270) and 72sin(270) but i got something like 43.2 instead of 43.6 if somebody could explain it to me it would help. thank you
 
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Did you draw a sketch? You can add the velocity of the airplane and the wind velocity (as vectors) to get the effective velocity and its direction. You can use that diagram to get the correct formula.
 
the problem is tho that you only have the velocity of the plane no? because you do not have its direction that is what you are trying to get the direction it should go in because while its supposed to be at a 38 angle the south wind is changing its direction. so when i tried doing it that way i did
580cos(38)=457.0462371
580sin(38) + 720=429.0836557
and if you then do arctan you get 43.2 which is not correct because it an angle of 43.6
while you can use the law of sines i am trying how to figure it out using the vector components
 
580 cos(38) etc. cannot be interesting because the 580 is the plane's speed relative to the air but the 38 is not the plane's heading relative to the air. Put in an unknown for the relative heading and obtain some equations.
 
so would the equation look something like this
580^2=(72+xsin(38))^2+(xcos(38))^2? if so how would one go abut solving it because i am having trouble figuring this out.
 
Actually, I said to introduce an unknown for the heading. That's an angle. But you've introduced x as the net speed, which is fine. And your equation looks right.
Can you not expand the squarings on the RHS to obtain a quadratic in x?
 
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