Question

A 33-N force acing due north and a 44-N force acting at 30 degrees act concurrently on point P. What is the magnitude and direction of a third force that produces equilibrium at point P?

Answer 1

"..acting at 30 degrees" is not clear.
Which way does the 30° go, east of north ?, west of north ?,

I will solve it as if it was 30° east of north. If not just make the necessary adjustments.

the resultant of our two forces, I will consider north to be 90°
= (33cos90, 33sin90) + (44cos60, 44sin60)
= (0, 33) + (22, 22√3)
= (22, 33+22√3)

magnitude = √(22^2 + (33+22√3)^2
= √( 484 + 1089 + 1452√3 + 1452)
= √(3025 + 1452√3)
= appr 74.43 N

angle of resultant:
tanØ = 1452√3/3025
Ø = appr 39.74°
So the force keeping it in equilibrium has to go in the opposite direction, which is 219.74° with a magnitude of 74.32 N
I will leave it to you to express the direction in the form needed. Make a sketch to determine it.

Answer 2

30 degrees which way? I assume to the East.

So you want the resultant, then the equilibrant will be opposite (read add 180 degrees).

Ok, forces: N direction
33 + 44cos30= you do it, N component

E direction:
44 sin30= you do it, E component

Now, the resultant:
magnitude: sqrt(N^2 + E^2)
direction: arctan (E/N) E of N

Equilibrate: same magnitude as resultant. Direction, add 180
so direction=arctan(E/N) W of S