My solution was that I would need to sum the $x$ and $y$ components for the net force. However, the solution manual say that the net force should be calculated by using the Pythogorean theorem. I believe am missing something as I don't really get why we need to do that, so please help me with this. I am attaching screenshots of the problem and the solution.
Since force is a vector, then technically, the net force is found by summing the vectors component-wise: $$\vec F_{\text{net}}=\vec F_1+\vec F_2=(F_{1x}+F_{2x})\mathbf{\hat i}+(F_{1y}+F_{2y})\mathbf{\hat j}.$$ However, to find the magnitude of the net force then one must seek to find the length of the force vector. The Pythagorean theorem is handy for that because the force vector forms the hypotenuse of a right triangle with respect to the $x,y$ axis. So that for the magnitude of a force vector $\vec F$, we calculate: $$|\vec F|=\sqrt{F_x^2+F_y^2}.$$ If the vector $\vec F$ is a sum of two other vectors then one can extrapolate from the above that: $$|\vec F_1+\vec F_2|=\sqrt{(F_{1x}+F_{2x})^2+(F_{1y}+F_{2y})^2}.$$ Now you know how to add two vectors and find the resultant magnitude, however, what about the direction? To do this specify the angle the force vector makes with respect to the x-axis. To do this take advantage of the definition of the tangent function $\tan\theta=F_y/F_x$, for a sum of vectors use the summed components: $$\tan\theta={(F_{1y}+F_{2y})\over (F_{1x}+F_{2x})}.$$ The angle $\theta$ is of course found from taking the inverse tangent function denoted $\tan^{-1}$ or $\arctan$. $$\theta=\tan^{-1}\bigg({(F_{1y}+F_{2y})\over (F_{1x}+F_{2x})}\bigg).$$ If you can write the vector in terms of its components, calculate its magnitude and find it's angle with respect to the x-axis, then you know pretty much what there is to know about it.
