Vectors problem, Please Help! A buoy is floating in water and is tied to a post. The water is creating a force of $3$ N on a bearing of $125^\circ$ and the wind is creating a force of $2$ N on a bearing of $230^\circ$
What force does the post need to exert on the buoy to keep it in equilibrium?
I have this, Is this the answer?
Bearing is the angle that is clockwise from north. To make this simpler, let’s assume that north is the positive $y$ direction and east is the positive $x$ direction. Since the first force is $125^\circ$ clockwise from north, this force is $35^\circ$ below the positive $x$ axis. This means its $y$ component is negative and its $x$ component is positive. Let’s use the following equations to determine these numbers. 
$$y = -F \cdot \sin \theta = -3 \cdot \sin 35 $$
$$x = F \cdot \cos \theta = 3 \cdot \cos 35 $$
To determine the angle, I subtracted $90^\circ$. The direction of this force is $35$˚south of east. 
Since the second force is $230^\circ$ clockwise from north, this force is $50$˚ clockwise from the negative $y$ axis. This means its $y$ component is negative and its $x$ component is negative. Let’s use the following equations to determine these numbers. 
$$y = -F \cdot \cos \theta = -2 \cdot \cos 50 $$
$$x = F \cdot \sin \theta = -2 \cdot \sin 50 $$
To determine the angle, I subtracted $180^\circ$. The direction of this force is $50^\circ$ west of south. Let’s add these components. 
$$y = --3 \cdot \sin 35 + -2 \cdot \cos 50 = -3.006304528 \mathrm{\ N} $$
$$x = 3 \cdot \cos 35 - 2 \cdot \sin 50= 0.925367247 \mathrm{\ N} $$
To determine the magnitude of the force, use the following equation. 
$$F = \sqrt{y^2 + x^2} = \sqrt{(-3.006304528)^2 + 0.925367247^2} = 3.145252417 \mathrm{\ N}$$
 A: I think instead of $--3$ you meant to just put $-3$
(especially since you then computed the correct result assuming
the value of that factor was actually $-3$).
Basically the approach is sound, and other than the stray $-$ symbol all
the intermediate results and the final result look good.
I tend to take a different approach
when programming a computer to do this sort of calculation,
because I would like to avoid the extra logic of having to figure out what
quadrant the vector is in and then subtract some multiple of $90^\circ$
from the angle. Assuming the direction is clockwise from the positive $y$ axis
(anathema to pure mathematics, but in practical navigation exercises I end
up doing this a lot),
I would write
$$ F_x = F \sin \theta, $$
$$ F_y = F \cos \theta. $$
I would then assume the $\sin$ and $\cos$ functions are evaluated correctly,
meaning they are positive when they should be positive and negative when they
should be negative, depending on which quadrant the angle happens to fall in.
In the end, the methods should give the same result, because for example
$3 \cos 125^\circ = -3 \sin 35^\circ,$ as can easily be confirmed
by the same methods you used.
