Trigonometry

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I need help:
If vector u has a magnitude of 4 meters and a direction of 17 degrees and vector v has a magnitude of 6 meters and a direction of 133 degrees. Find the magnitude of the resultant vector of u + 3v.

  • Trigonometry -

    I made a sketch
    labeled end of first trip A, end 2nd trip B to get triangle OAB
    OA=4 , AB = 18 and angle BAO = 64° by simple geometry
    OA^2 = 18^2+4^2 - 2(4)(8)cos64 , by cosine law
    ..
    OA = 16.6395 m

    by sine law: Let angle AOB= Ø
    sinØ/18 = sin64/16.6395
    ...
    Ø = 76.5 or 180-76.5
    Ø = 76.5 or 103.5
    looking at my sketch, I chose 103.5°
    so the new direction angle is 103.5+17= 120.5°

    Maginitude is 16.6395 m and direction is 120.5°

    or

    u = (4cos17 , 4sin17)
    3v = (18cos133, 18sin133)
    u + 3v = (4cos17 , 4sin17) + (18cos133, 18sin133)
    = (-8.45 , 14.338)
    maginitude = √( (-8.45)^2 , 14.338^2 ) = 16.6395

    direction angle
    tanØ = 14.338/-8.45
    Ø = 120.5°

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