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Math Help - [SOLVED] Problem having to do with bearing and planes

  1. #1
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    [SOLVED] Problem having to do with bearing and planes

    Hi Everybody!

    I have this math problem and I solved it but I'm not sure if I solved it correctly as finding the angle of the bearing always messes me up!:

    A Boeing 747 Jumbo Jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant 80 mph from the direction of 260 degrees. Find the actual speed and direction of the plane.

    My answers:
    Airplane Speed: 617.24 mph and the angle at which it is flying is 49.345 degrees.

    Any help is appreciated,
    Stealth
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  2. #2
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    Lexington, MA (USA)
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    Hello, Stealth!

    Sorry, your answers are off . . .


    A Boeing 747 Jumbo Jet maintains an airspeed of 550 mph in a SW direction.
    The velocity of the jet stream is a constant 80 mph from the direction of 260.
    Find the actual speed and direction of the plane.
    Code:
                        N
                        :
                        :
      W - - - - - - - - oA 
            C o       * :
                    *45:
                  *     :
                *       :
              *         :
            *           :
          *             :
        *               :
      o                 :
      B                 S

    Draw segments AC and BC.
    . . (Hope you can follow my labeling.)

    The plane is flying from A to B: . AB = 550,\;\angle BAS = 45^o

    The wind blows from C to A: . CA = 80
    Major angle NAC = 260^o\quad\Rightarrow\quad \angle WAC = 10^o \quad\Rightarrow\quad \angle CAB = 35^o


    In \Delta CAB\!:\;CB^2 \:=\:CA^2 + AB^2 - 2(CA)(AB)\cos(\angle CAB)

    . . . . . . . . CB^2\:=\: 80^2 + 550^2 - 2(80)(55)\cos35^o \:=\:236814.6201

    Therefore: . \boxed{CB \;\approx\;486.6\text{ mph}}


    We have: . \cos(\angle ACB) \:=\:\frac{CA^2 + CB^2 - AB^2}{2(CA)(CB)} \:=\:\frac{80^2+486.6^2 - 550^2}{2(80)(486.6)} \:=\:-0.761925092

    Hence: . \angle ACB \;\approx\;139.6^o

    Therefore, the bearing of CB is about: . \boxed{219.6^o}

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  3. #3
    Newbie
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    Oh ok. I see what you did. But than what did I do wrong?:

    I tried to imagine a picture as you did, but than I solved it like this:

    V1 = 550(cos(225), cos(225))
    V2 = 80(cos(190), cos(190))

    V1 + V2 = (-467.6933499, -402.8005839)

    squareroot(V1squared + V2squared) = 617.24 mph

    and than for the angle I did it like this:

    Tanx = -402.800539/-467.6933499 = arctan(1.16) = 49.345 degrees.

    Can you explain what I may have done wrong because I'm fairly sure my set up is correct.

    Thanks,
    Stealth
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  4. #4
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    Never mind got it! Added instead of subtracted!

    Thanks sooo much!

    Stealth
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