Asked by Annie
In still air, an airplane is flying at 100 mph. It encounters a wind blowing toward the west at 50 mph. What should be the plane's compass heading for its course to be N30ºW?
Answers
Answered by
Steve
If the plane is flying on heading NθE, then we want
100sinθ-50 = -v/2
100cosθ = v √3/2
100sinθ = 50-v/2
100cosθ = v√3/2
(50-v/2)^2 + 3/4 v^2 = 100^2
v = 115.14
So, cosθ = .997
θ = 4.3º
and the heading is N4.3ºE
100sinθ-50 = -v/2
100cosθ = v √3/2
100sinθ = 50-v/2
100cosθ = v√3/2
(50-v/2)^2 + 3/4 v^2 = 100^2
v = 115.14
So, cosθ = .997
θ = 4.3º
and the heading is N4.3ºE
Answered by
Annie
Could you explain the (50-v/2)^2 + 3/4v^2 = 100^2 part? I'm a little confused.
Answered by
Steve
square the previous equations and add them up. Recall that
sin^2θ + cos^2θ = 1
sin^2θ + cos^2θ = 1
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