If I were to sketch the triangle for this problem on the cartesian plane, it would be in the fourth quadrant (since the side opposite to $\theta$ is negative).
From the values of the opposite and adjacent side, I know that the triangle is a $30^\circ$, $60^\circ$, $90^\circ$ triangle of which the sides are $1$, $\sqrt3$ and $2$.
With this information, how can I calculate the angles whose cotangents are equal to $-\frac{1}{\sqrt3}$ without the use of a calculator?

  • $\begingroup$ Using the fact that:1) $\cot(-x)=-\cot x$ 2) $\cot y=\cot x\iff x-y=k\times180^\circ, k\in\Bbb Z$. Together with a sensible consideration of the cotangents of the angles of the triangle you've been given. $\endgroup$ – user228113 Jun 12 '16 at 10:03


Now, it can be either

$$\tan\theta=-\sqrt3=\frac{-\frac{\sqrt3}2}{\frac12}=\frac{\sin\left(-\frac\pi3\right)}{\cos\left(\frac\pi3\right)}\implies\theta=-\frac\pi3+k\pi\;,\;\;k\in\Bbb Z$$

or else

$$\tan\theta=-\sqrt3=\frac{\frac{\sqrt3}2}{-\frac12}=\frac{\sin\left(\frac{2\pi}3\right)}{\cos\left(\frac{2\pi}3\right)}\implies\theta=\frac{2\pi}3+k\pi\;,\;\;k\in\Bbb Z$$

Of course, you get the very same solution set ( observe, for example, that $\;-\frac\pi3=\frac{2\pi}3-\pi\;$) , but it is perhaps a little easier to get the final solution sets for your specific problem in the second form, since you need positive values of the angle:



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