Jtdylktuy

When finding the direction angle with the formula $theta = tan^{-1} left(frac{y}{x}right )$, when do you add $180$ degrees to the answer? Is it whenever the $x$ is negative, when the angle is in the third or fourth quadrant (if this is the case, how would I know the angle is there?), or just in the third quadrant (if this is the case, how would I know the angle is there)? Or something else entirely?

If you look at this table in the Wikipedia article, you will see that in the arctangent row, the "range of usual principal value" is given as $-90^circ<y<90^circ$.

This means that if you give a number to the arctangent function, most calculators respond with an answer between $-90^circ$ and $90^circ$. This is the half-plane on the right, quadrants I and IV, so $x$ is assumed positive.

If $x$ is negative, the answer you want is $180^circ$ away.

Use the following formula and no worry to add something.

f(x,y)=180-90*(1+sign(x))* (1-sign(y^2))-45*(2+sign(x))*sign(y)

  -180/pi()*sign(x*y)*atan((abs(x)-abs(y))/(abs(x)+abs(y)))

Andrew Woods' answer is correct, but let me offer another way to compute $theta$.

As shown in this answer,
$$
begin{align}
tan(theta/2)
&=frac{sin(theta)}{1+cos(theta)}\
&=frac{frac yr}{1+frac xr}\[3pt]
&=frac{y}{x+r}
end{align}
$$
which leads to the formula
$$
bbox[5px,border:2px solid #C0A000]{theta=2arctanleft(vcenter{frac y{x+sqrt{x^2+y^2}}}right)}
$$

which is valid as long as $x+sqrt{x^2+y^2}ne0$; that is, $yne0$ or $xgt0$.

Verification

Since $tan(2x)=frac{2tan(x)}{1-tan^2(x)}$,
$$
begin{align}
tan(theta)
&=frac{2frac y{x+sqrt{x^2+y^2}}}{1-left(frac y{x+sqrt{x^2+y^2}}right)^2}\
&=frac{2yleft(x+sqrt{x^2+y^2}right)}{2x^2+2xsqrt{x^2+y^2}}\[12pt]
&=frac yx
end{align}
$$