Asked by Morgan
The problem is: an airplane is flying in the direction 120 degrees with an airspeed of 320 mph, and a 35 mph wind is blowing in the direction 60 degrees. The question ask to find the true course and the ground speed. To solve the problem I have done this:
Flying 120sin 320_+120cos320_
Wind 60sin(35-180)_+60cos(35-180)_
I understand that I then add the together. The problem is where there are _ I don't know what direction I am suppost to put and how to determine it.
Flying 120sin 320_+120cos320_
Wind 60sin(35-180)_+60cos(35-180)_
I understand that I then add the together. The problem is where there are _ I don't know what direction I am suppost to put and how to determine it.
Answers
Answered by
Reiny
(320cos120 , 320sin120) + (35cos60 , 35sin60)
you should be familiar with the trig ratios of the 30-60-90 triangle, and the multiples of those angles using the CAST rule
= (320(-1/2) , 320(√3/2) + (35(1/2) + 35(√3/2)
= (-160 , 160√3) + (35/2 , 35√3/2)
=( -142.5 , 177.5√3)
magnitude = √( (-142.5)^2 + (177.5√3)^2 )
= appr 338.9 mph
direction:
tanØ = 177.5√3/-142.5 = -2.157...
Ø = 180 - 65.13 = appr 114.9°
I you make a sketch you will see that you could also use the cosine law,
I got the same answer that way.
you should be familiar with the trig ratios of the 30-60-90 triangle, and the multiples of those angles using the CAST rule
= (320(-1/2) , 320(√3/2) + (35(1/2) + 35(√3/2)
= (-160 , 160√3) + (35/2 , 35√3/2)
=( -142.5 , 177.5√3)
magnitude = √( (-142.5)^2 + (177.5√3)^2 )
= appr 338.9 mph
direction:
tanØ = 177.5√3/-142.5 = -2.157...
Ø = 180 - 65.13 = appr 114.9°
I you make a sketch you will see that you could also use the cosine law,
I got the same answer that way.
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