Plz help..Two vectors A and B have values 27N and 61N, respectively.?

If vector A is at an angle of 51 degrees and vector B is at an angle of 84 degrees.What is the vector difference of the two vectors?

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  • Vaman
    Lv 7
    4 weeks ago

    Magnitude remain the same.=27+61=88. Now you need to find the resulting angle. The angle betweem A and B is 84-51=33. The magnitude of  the resultant will be

    (vec A-vec B)^2= A^2+B^2-2AB cos 33=r^2

    r^2=1687.4 r=41

    The angle of the resultant is given by

    27= 41 cos a, cos a=27/41  a=48.81degrees

    wrt x axis the resultant makes an angle 51+48.81=99.81.

  • Ax = 27 * cos(51)

    Ay = 27 * sin(51)

    Bx = 61 * cos(84)

    By = 61 * sin(84)

    A - B

    Ax - Bx = 27 * cos(51) - 61 * cos(84)

    Ay - By = 27 * sin(51) - 61 * sin(84)

    m = sqrt((Ax - Bx)^2 + (Ay - By)^2)

    m^2 =>

    (Ax - Bx)^2 + (Ay - By)^2 =>

    (27 * cos(51) - 61 * cos(84))^2 + (27 * sin(51) - 61 * sin(84))^2 =>

    729 * cos(51)^2 - 3294 * cos(51) * cos(84) + 3721 * cos(84)^2 + 729 * sin(51)^2 - 3294 * sin(51) * sin(84) + 3721 * sin(84)^2 =>

    729 * (cos(51)^2 + sin(51)^2) + 3721 * (cos(84)^2 + sin(84)^2) - 3294 * (cos(84) * cos(51) + sin(84) * sin(51)) =>

    729 * 1 + 3721 * 1 - 3294 * cos(84 - 51) =>

    729 + 3721 - 3294 * cos(33) =>

    4450 - 3294 * cos(33)

    m = sqrt(4450 - 3294 * cos(33))

    m * cos(t) = Ax - Bx

    m * cos(t) = 27 * cos(51) - 61 * cos(84))

    cos(t) = (27 * cos(51) - 61 * cos(84)) / sqrt(4450 - 3294 * cos(33))

    t = +/- 75 degrees

    m * sin(t) = Ay - By

    sin(t) = (27 * sin(51) - 61 * sin(84)) / sqrt(4450 - 3294 * cos(33))

    t = ‭-75 , ‭255

    t = -75

    m = ‭41.078207716352149123906185623641‬

    41 Newtons at -75 degrees

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