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March 29, 2017

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This has to do with vectors...I'm wondering if I set up the diagram correctly?

A plane has a speed of 250mph and flies with a bearing of N35degE. The wind is blowing from the west at 25 mph with a heading of due east. What is the actual speed and heading of the plane?

  • Trig - ,

    Whoops I forgot to include the diagram...here it is:

    i.imgur [dot] com/RRY5pox.png

  • Trig - ,

    I would tack on the wind vector to the end of the plane vector at the point P so the final end point is Q
    Join Q to the origin O

    vector OQ has our required direction and its magnitude represents the actual speed
    By simple geometry angle P = 125°

    so by the cosine law:
    OP^2 = 250^2 + 25^2 - 2(250)(25)cos 125°
    = 70294.705..
    OP = 265.13
    So the actual speed is 265.13 mph

    Let the additional angle be Ø
    by the sine law:
    sinØ/25 = sin 125/265.13
    sinØ = .07724
    Ø = 4.43°
    Ø + 35° = 39.43°
    So the heading should be N 39.43° E

  • Trig - ,

    Thank you <3 you are a lifesaver!

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