An object in equlibrium has three forces acting on it. A 33-N force
acts at 90 degrees from the x-axis and a 44-N force acts at 60
degrees.
What are the magnitude and direction of the third force?

This is a problem slightly more difficult than resolving three forces.

First, I will resolve the two known forces, then the third force will
be the opposite of that.

To resolve two forces, one must first take the horizontal and vertical
components of each force, then add them. The 33N force (force A), is
easy. It is only a vertical force. Use trigonometry to find the 44N
force (force B). I find it easier to make a table like this one to
help me out:

A   0    33
B  22 38.10

You then add them together, and then use Pythagorus to merge them.
(This is triangle, remember.)

sqrt (22^2 + 71.1^2) = 74.42N.

That is the magnitude of the resultant force. Vectors have direction
and magnitude. Force is a vector. The force is easy. Our triangle that
we made with our resultant force has an angle. Because tan is opposite
over adjacent, and we want to find the displacement from x-axis, 71.2
is our adjacent, and  22 is our opposite. That means our resultant is
17.19 degrees south from the x-axis.

(email me if I'm right: huw.rowland@gmail.com)