Solve for Radian Exactly $$\tan(A) = \frac{\sqrt{3}}{-3}$$
I've tried using special triangles but couldn't find a matching faction using sohcahtoa.
 A: $\tan(A) = \frac{\sqrt{3}}{-3} = \tan(A) = \frac{-1}{\sqrt3}$ ... (1)
Note that $\tan(A) = \frac{\sqrt 3}{-3}$ is also $\frac { 1}{-\sqrt 3}$ ... (2)
That is, the angle A could be in the 2nd [as in case (1)]or the 4th quadrant [as in case (2)].
A: The definition “tangent equals opposite over adjacent” is valid only for angles $\theta$ that satisfy $0<\theta<90^\circ$. For other angles, you have to use a definition that’s based on analytic geometry. Haven’t you seen it? It goes like this: you have an angle $\theta$, it can be positive or negative, of any size. You draw a (half-)line outward from the origin in the $(x,y)$-plane that makes an angle of $\theta$ with the positive $x$-axis: if $\theta>0$, you turn counterclockwise by an amount of $\theta$, if negative, you turn clockwise by an amount of $|\theta|$. For example, if your angle is $135^\circ$, your “ray” out from the origin is in the second quadrant, bisecting the angle between the positive $y$-axis and the negative $x$-axis.
Now take any point $(x,y)$ on your just-constructed ray. Then the tangent of $\theta$ is $y/x$. It’s this that’s the appropriate definition of tangent. In our example, the tangent is $-1$.
A: Consider the "special right triangle" that has angles of $\frac{\pi}{2}, \frac{\pi}{3}, \frac{\pi}{6}$ and sides of $1, \sqrt 3, 2$. The $2$ is the length of the hypotenuse. Verify that Pythagoras' Theorem is satisfied.
Now what angle has the tangent $\frac{1}{\sqrt{3}}$? To help you see this, remember that the two sides at right angles to each other measure $1$ and $\sqrt 3$. There are two angles apart from the right angle and you already know the measure of these two angles (in my previous paragraph). Remember that the greater angle will lie opposite the longer side.
So let's say you've found that angle and let's call that angle $x$. This is the first quadrant value that has a tangent of the same magnitude (regardless of sign) as the angle $A$ that you want to find, i.e. $|\tan A| = |\tan x| = \frac{1}{\sqrt 3}$
Once you find $x$, you need to remember that tangent is negative in the second and fourth quadrants. So there are two possible values for $A$. The second quadrant value can be calculated as $A = \pi - x$ and the fourth quadrant value can be represented either as $A = 2\pi - x$ or simply, $A = -x$, depending on the range you're working within. Remember that angles are basically periodic so two values can be equivalent ways of representing the same angle.
