Finding Air and Ground Speed Using Trigonometry: A Pilot's Navigation Problem

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A pilot needs to determine airspeed and groundspeed while flying on a bearing of 68.2 degrees, with a wind blowing from the south at 38 MPH. The initial confusion arose from the diagram setup, as the pilot was unsure how to represent the relationship between heading, true course, and wind direction. Clarifications indicated that the plane's heading is due east, while the true course is at 68.2 degrees, forming a right triangle for calculations. After some discussion, the pilot realized the correct approach to drawing the necessary diagram. The problem was ultimately resolved with the right visual representation.
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Homework Statement


A pilot wants to fly on a bearing of 68.2 degrees. By flying due east, he finds that 38 MPH, blowing from the south, puts him on the course. Find the air speed and ground speed.

Homework Equations


using law of sines/cosines, ambiguous case.

The Attempt at a Solution


Tried drawing a diagram according to the book, the one with grspeed/airspeed/wind direction/heading.

however, it seems like it's moving south and i would have to draw a different diagram to solve this.

i have NO idea where to start! this has stumped me. I'm not sure if i even have enough information to complete this problem. i just need a general idea of what kind of diagram i need to draw..
 
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clook said:
however, it seems like it's moving south and i would have to draw a different diagram to solve this.

The plane isn't moving south. The question states that, by flying due east, the wind (blowing from the south) ensures that the plane travels on the correct path of 68.2 degrees. How have you set up the diagram?

For the second part of the question, I don't know what the groundspeed means.
 
http://savemyfile.net//files/6/book.jpg

something similar to this
 
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Ahh, ok I get what you mean now! Ok, well try drawing a diagram similar to that. The heading is the direction in which the pilot flies, ie due east here. The true course is the direction the pilot wants to fly, ie 68.2 degrees. The wind is blowing from the south. Your triangle will be different to the one in the book, in that it will be a right angled triangle.
 
argh.. it would bea right triangle?

man, is there a way you could provide me some type of diagram for this?

edit: nvm, got it.
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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