Resolving runs both ways. Given two forces, put each one's shadows on the axes, add the x-shadows to each other and the y-shadows to each other — components add like ordinary numbers, because they lie on the same line — and you are left with one $F_x$ and one $F_y$ describing the whole crowd. That single arrow is the resultant: $|\vec{F}| = \sqrt{F_x^2 + F_y^2}$, read aloud the magnitude equals the root of F-x squared plus F-y squared, and it points along $\theta = \tan^{-1}\!\left(\dfrac{F_y}{F_x}\right)$ — theta equals inverse tan of F-y over F-x. Both components are in newtons, the resultant is in newtons, and $\theta$ comes out in degrees measured from the $+x$ axis.

The one thing that never works is adding the MAGNITUDES. A 30 N pull east and a 40 N pull north do not make 70 N; they make 50 N, at 53° north of east. If they did add, walking two blocks east and two north would be the same as walking four blocks east, and no city has ever worked that way.

When the two forces meet at some angle $\theta$ that is not a right angle, the parallelogram rule handles it in one line: $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$, where $A$ and $B$ are the two magnitudes in newtons and $\theta$ is the angle BETWEEN them, not either one's bearing. Set $\theta = 90^\circ$, watch $\cos 90^\circ = 0$ delete the last term, and Pythagoras walks back out. One formula, two faces.