Finding $(+) (-)$ direction with Cosine and Sine In the kind of exercise where they tell us to find the resultant force $F = F_1 + F_2$. We can know the magnitude by calculating the $x-$ and $y-$ components, but how do we know the direction of the resultant force (as the magnitude of the resultant force is as below, where there is no cosine or sine to show whether the direction is negative or positive). 
I was able to find:


*

*The direction of the $x-$ and $y-$ component vectors' directions.

*The magnitude of the resultant force.
But I fail to find the resultant force's direction.


 A: The first step is to fix a coordinate system. You can chose any coordinate system you want, but if you pick one that is convenient, calculations become much simpler. Since here you have the angles given to the horizontal, it is natural to pick one coordinate (let's say $x$) along the horizontal and for convention let it point to the right. Let's put the other coordinate ($y$) be perpendicular up and the origin where the forces meet.
Adding  forces is very easy, just like vectors you just need to add the components of the forces:
$$\vec{F}=\vec{F}_1+\vec{F}_2=\begin{pmatrix}F_{1x}\\F_{1y}\\F_{1z}\end{pmatrix}+\begin{pmatrix}F_{2x}\\F_{2y}\\F_{2z}\end{pmatrix}$$
This works in all coordinate systems. Specifically in your case you have all forces in a 2D plane, so you can ignore the third ($z$) coordinates.
To find the components of the forces you can either: 
(1) always take the angle to the positive $x$ direction ($\phi_{x+}$) and then you don't need to worry about any minus signs as the coordinates will be $F_x=F\cos\phi_{x+}$ and $F_y=F\sin\phi_{x+}$. In your case that would be for instance $F_{1x}=F_1\cos (180^\circ-45^\circ)$ and $F_{1y}=F_1\sin (180^\circ-45^\circ)$
or (2) take the projection to any axis and add a minus sign if that projection points in the negative direction of the axes. In your case the projection of $\vec{F}_1$ to the x-axis would be $F_1\cos(45^\circ)$. From the figure you see that this points in the negative direction so the vector component will get an additional minus sign: $F_{1x}=-F_1\cos(45^\circ)$.
Once you have done all this, you should have $\vec{F}_1$ and $\vec{F}_2$ written in coordinates, you can add these coordinates to obtain the coordinates of the resultant force, $(F_x, F_y)$.
In the same way that you expressed the components of $\vec{F}_1$ and $\vec{F}_2$ in terms of the angle, you now have to reverse this procedure in order to get the angle from the components. Using the first method outlined above you know that the relationship between the angle to the positive x-direction and the coordinates is:
$$F_x=F\cos\theta\\
F_y=F\sin\theta$$
or alternatively:
$$\tan\theta=\frac{F_y}{F_x}$$
where you have to be careful with special cases where $\cos\theta=0$
Since you know the coordinates of the resultant force, you can find the angle from these equations easily.
A: The first step is to break the F1 and F2 vectors into their x and y components.  If an x-component is applied west, it is negative (same with a y-component being applied south, it would be negative then).
Then, add up the two x-components and the two y-components.  This leaves you with one net y-component and one net x-component.
At this point you can form a right triangle with the x-component and y-component.  This hypotenuse is the resultant vector.  You can figure out its magnitude by using the Pythagorean Theorem.  You can figure out the angle between the resultant vector and the x-axis the cosine ratio:
cos theta = adjacent / hypotenuse = x-component / resultant vector
theta = cos^-1 (x-component / resultant vector)
Then just plug in your values to solve for theta.  If you wanted to find the angle between the resultant vector and y-axis, you would do:
cos theta = adjacent / hypotenuse = y-component / resultant vector
theta = cos^-1 (y-component / resultant vector)
Now you have the angle of your resultant vector either relative to the x-axis or y-axis.
A: Use the formula:
$$\alpha=\tan^{-1}\left[\dfrac{F_{2}\sin\theta}{F_{1}+F_{2}\cos\theta} \right]$$
where:
$\alpha=$ angle of resultant vector w.r.t. vector $F_{1}$
$\theta=$ angle between $F_{1}$ and $F_{2}$
