The vector A is 12 units long and points 30°north of east, The other vector B is 18 units long and points 45°west of north.?

 Find a vector C such that A+B+C=0  

4 Answers

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  • Anonymous
    3 weeks ago

    Ax = 12*cos 30 = 6√3

    Ay = 12*sin 30 = 6.0

    Bx = -18*cos 45 = -9√2

    By = 18*sin 45 = 9√2

    X = 6-9√2

    Y = 6√3+9√2

    Cy = -(6√3+9√2)

    Cx = 9√2-6

    module of C = √X^2+Y^2 = √(6-9√2)^2+(6√3+9√2)^2 = 24.08

  • JOHN
    Lv 7
    3 weeks ago

    Answer is below

    Attachment image
  • Vaman
    Lv 7
    3 weeks ago

    C is the resultant of A and B. Let us East is the direction. Now it is at 30 to x axis. B is at 45 degrees. Its angle from x axis will be 90+45=135. The angle difference is 135-30=105.

    The resultant r^2 will be

    A^2+B^2 + 2AB cos theta= 12^2+18^2+ 2*12*18 cos 105=356.19, r=18.9

    The angle between A and r is

    12= 18.9 cos a, cos a=12/18.9 a=50.56. This is the solution of equilibrium for C from the end ob at and angle 50.56 to A.

  • 12 * cos(30) + 18 * cos(90 + 45) + r * cos(t) = 0

    12 * sin(30) + 18 * sin(90 + 45) + r * sin(t) = 0

    12 * (sqrt(3)/2) + 18 * (-sqrt(2)/2) + r * cos(t) = 0

    12 * (1/2) + 18 * (sqrt(2)/2) + r * sin(t) = 0

    6 * sqrt(3) - 9 * sqrt(2) + r * cos(t) = 0

    6 + 9 * sqrt(2) + r * sin(t) = 0

    r * cos(t) = 9 * sqrt(2) - 6 * sqrt(3)

    r * sin(t) = -(9 * sqrt(2) + 6)

    r * sin(t) / (r * cos(t)) = (9 * sqrt(2) - 6 * sqrt(3)) / (-(9 * sqrt(2) + 6))

    tan(t) = (6 * sqrt(3) - 9 * sqrt(2)) / (6 + 9 * sqrt(2))

    tan(t) = (6 * sqrt(3) - 9 * sqrt(2)) * (6 - 9 * sqrt(2)) / (36 - 18)

    tan(t) = 3 * (2 * sqrt(3) - 3 * sqrt(2)) * 3 * (2 - 3 * sqrt(2)) / 18

    tan(t) = 9 * (4 * sqrt(3) - 6 * sqrt(6) - 6 * sqrt(2) + 18) / 18

    tan(t) = (4 * sqrt(3) - 6 * sqrt(6) - 6 * sqrt(2) + 18) / 2

    tan(t) = 2 * sqrt(3) - 3 * sqrt(6) - 3 * sqrt(2) + 9

    t = tan-1(2 * sqrt(3) - 3 * sqrt(6) - 3 * sqrt(2) + 9)

    t = ‭41.120689501598006426067296322752‬....

    r^2 * sin(t)^2 + r^2 * cos(t)^2 = (9 * sqrt(2) - 6 * sqrt(3))^2 + (-(9 * sqrt(2) + 6))^2

    r^2 * (sin(t)^2 + cos(t)^2) = 81 * 2 - 108 * sqrt(6) + 36 * 3 + 81 * 2 + 108 * sqrt(2) + 36

    r^2 * 1 = 162 + 108 + 162 + 36 - 108 * sqrt(6) + 108 * sqrt(2)

    r^2 = 324 + 144 + 108 * (sqrt(2) - sqrt(6))

    r^2 = 468 + 108 * (sqrt(2) - sqrt(6))

    r^2 = 36 * (13 + 3 * (sqrt(2) - sqrt(6)))

    r = 6 * sqrt(13 + 3 * (sqrt(2) - sqrt(6)))

    r = ‭18.873001152856188777297824935883‬

    19 units at 41 degrees North of East ought to get you close.

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