What does it mean by a force acting horizontally? Question:
A body of $10kg-wt$ is kept in an equilibrium condition by a force of $5kg-wt$ acting horizontally, a force of $F\ kg-wt$ acting at an angle $60^{\circ}$ with the horizon and a force of $R\ kg-wt$ acting perpendicular to $F$. Show that $F-\sqrt{3}R+10=0$.
My book's attempt:

$$5\cos(0^{\circ})+10\cos(-90^{\circ})+F\cos(60^{\circ})+R\cos(150^{\circ})=0$$
$$5+0+\frac{F}{2}-\frac{\sqrt{3}R}{2}=0$$
$$10+F-\sqrt{3}R=0$$
$$F-\sqrt{3}R+10=0\ \text{(showed)}$$
My attempt:

$$5\cos(0^{\circ})+10\cos(-90^{\circ})+R\cos(30^{\circ})+F\cos(120^{\circ})=0$$
$$5+0+\frac{\sqrt{3}R}{2}-\frac{F}{2}=0$$
$$10+\sqrt{3}R-F=0$$
My comments:
I was not able to prove the statement given in the question while my book was able to. This is because my figure is different than that of my book, but I believe that I didn't do anything outside the capacity of the question. I think I drew the figure according to the question. The difference between my attempt and my book's attempt is that I drew the horizontal force acting towards the left while my book drew it acting towards the right. Both should be correct, according to the question, since the book didn't specify towards which side, left or right, the horizontal force was acting on. Then why was I unable to prove the statement in the question?
My question:

*

*Why couldn't I prove the question statement correctly when I drew the figure according to the question?

 A: 
The difference between my attempt and my book's attempt is that I drew the horizontal force acting towards the left while my book drew it acting towards the right. Both should be correct, according to the question, since the book didn't specify towards which side, left or right, the horizontal force was acting on. Then why was I unable to prove the statement in the question?

Force is a vector, and implicitly has direction as well as magnitude.
A force of $5 \text{ kg m/s}^2$ horizontally implicitly has positive direction, because $5>0$. If we center our coordinate system at whatever is being pushed, then positive direction is (by convention) to the right).
Similarly, a force of $-5 \text{ kg m/s}^2$, having negative direction, would be going left in this scenario.
In fact, the same could be said for angles; in this scenario, we measure the angle starting from the positive side of the $x$-axis. (Otherwise, an angle of $-60^\circ$ would still trace out the same angle, in terms of size, but be in the wrong quadrant.)
