Basic physics question - Vector nature of forces: Forces in two dimensions?

An object is being acted upon by three forces and moves with a constant velocity. One force is 60.0 N along the x-axis, the second is 75.0 N along a direction making a counterclockwise angle of 150.0° with the x-axis. What is the magnitude and direction of the third force, measured counterclockwise from the x-axis?

I assume that the angle of 150.0° with the x-axis means the 150.0° counterclockwise from the positive x-axis. This means the 2nd velocity vector has a negative x-component and a positive y-component

Since the object moves with a constant velocity, the acceleration = 0

So, the sum of the forces in the x and y directions = 0

Sum of x-components = F * cos θ + 60 + (75 * cos 150) = 0

F * cos θ + 60 + -64.95 = 0

F * cos θ = 4.95 = x component of F

Sum of y-components = (F * sin θ) + (75 * sin 150) = 0

F * sin θ = -37.5 = y-component of F

Since the x-component is positive and the y-component is negative, the force vector is in the quadrant with +x and –y.

F * sin θ ÷ F * cos θ = tan θ

tan θ = -37.5 ÷ 4.95

θ = tan^-1 (-37.5 ÷ 4.95) = -82.48°

This means the force vector is 82.48° clockwise from the positive x-axis, so the force vector is 360° – 82.48 = 277.52° counterclockwise from the positive x-axis.

F * sin -82.48° = -37.5

F = 37.82 N

F * cos -82.48° = 4.95

F = 37.82 N

3rd Force = 37.82 N at angle of 277.52° counterclockwise from the positive x-axis.

Beca1b5b72d19c9d374e131ecf012bc30e8se the cans are sliding down the ra1b5b72d19c9d374e131ecf012bc30e8p with consistent velocity it is persevering with acceleration the forces of li1b5b72d19c9d374e131ecf012bc30e8iting friction and the that d1b5b72d19c9d374e131ecf012bc30e8e to gravity 1b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e8st be balancing one yet another to 1b5b72d19c9d374e131ecf012bc30e8ake internet rigidity = 01b5b72d19c9d374e131ecf012bc30e8 Or ass1b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e8ing '1b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e8' to be the coefficient of friction between the s1b5b72d19c9d374e131ecf012bc30e8rfaces we've 1b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e8461b5b72d19c9d374e131ecf012bc30e8cos 1b5b72d19c9d374e131ecf012bc30e84 = 461b5b72d19c9d374e131ecf012bc30e8sin1b5b72d19c9d374e131ecf012bc30e84 or 1b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e8 = tan 1b5b72d19c9d374e131ecf012bc30e84 = 01b5b72d19c9d374e131ecf012bc30e84451b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e8 permit 's' be the gap the can slide on the sa1b5b72d19c9d374e131ecf012bc30e8e horizontal s1b5b72d19c9d374e131ecf012bc30e8rface1b5b72d19c9d374e131ecf012bc30e8 we've 31b5b72d19c9d374e131ecf012bc30e841b5b72d19c9d374e131ecf012bc30e8^2*mu*9.8 = 1b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e846/(40 six/1b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e8)1b5b72d19c9d374e131ecf012bc30e8s or s =(31b5b72d19c9d374e131ecf012bc30e841b5b72d19c9d374e131ecf012bc30e8^2*mu*9.8)/[1b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e81b5b72d19c9d374e131ecf012bc30e8] = 11b5b72d19c9d374e131ecf012bc30e834 2*mu*9.8