a certain force has magnitude of 200N, its x component is -80N. what is its y component and its direction? two possible answers exist ... reply quickly..

F = 200

Fcos@ = F_x    (x compponent ))        _{...............1}

Fsin@ = F_y      ( y component )         _{.............2}

from 1

200cos@ = -80

 cos@ = -2/5

here @ can be in 2nd or 3rd quadrant ,

if @ is in 2nd quadrant then cos@ is -ve, sin@ is +ve@

cos@ = -2/5 , sin@ = (1-cos²@)^1/2 = (21)^1/2/5

now , F_y = F sin@ = 200(21)^1/2/5 = 184N

if @ is in 3rd quadrant then cos@ is -ve , sin@ is -ve

cos@ = -2/5 , sin@ = -(21)^1/2/5

F_y = Fsin@ = -184N

therefore two possible answers are 184N & -184N

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