Aparticle undergoes two displacements. the first has a
magnitude of 150 cm and makes an angle of 120° with
the positive x axis. the resultant displacement has a magnitude
of 140 cm and is directed at an angle of 35.0° to
the positive x axis. find the magnitude and direction of
the second displacement.

First you need to draw the picture of the problem to better understand it. Like the one bellow. 

In this task you have 2 sides of triangle and we can calculate angle between them. Angle between them is 120 - 35 = 85 degrees.

Once you have those 3 variables you can calculate third side of triangle using cosine law.
a - second displacement
b - first displacement
c- resultant displacement.


now we just need to calculate this.

a = 196

now, we use cosine law again to find the angle between second and first displacement.
degrees

The angle marked with "?" in the graph is our direction angle. We will call it 



Second displacement has magnitude of 196 and a direction of -14.64 with positive x axis

Distance covered, s = 8ft - 4ft = 4 ftinitial velocity, u = 34 ft/sthe acrobat is moving upwards and therefore, a = -g = -9.81 m/s^2 = 32.19 ft/s^2applying equation of motion; s = ut -1/2gt^2then,4 = 34t -1/2*32.19t^24 = 34t - 16.095t^2rearranging; -16.095t^2 + 34t - 4 = 0solving the quadratic equation; t = {-34 +/- sqrt [34^2 - 4(-16.)]}/2(-16.095) = 1.06 +/- 0.93 = 0.13 or 1.99the acrobat will take 0.13 seconds to be at 8 ft above the ground. additionally, once at the maximum height, the acrobat will move down and be at the same spot (that is height) after 1.99 seconds moving down.