# How do you find θ, if 0 < θ < 360 and tan theta = sqrt3  and theta in in QIII?

May 12, 2015

In QIII $\sin \theta < 0$ and $\cos \theta < 0$.

Notice that:

$\sqrt{3} = \frac{- \frac{\sqrt{3}}{2}}{- \frac{1}{2}}$

$= \frac{- \sin 60}{- \cos 60}$

$= \sin \frac{180 + 60}{\cos} \left(180 + 60\right)$

$= \frac{\sin 240}{\cos 240}$

$= \tan 240$

To see that $\cos 60 = \frac{1}{2}$, picture an equilateral triangle with sides of length 1 and cut it in half to produce two right angled triangles with internal angles 30, 60 and 90 degrees. The length of the shortest side is 1/2, the length of the hypotenuse is 1 and the length of the other side will be:

$\sqrt{{1}^{2} - {\left(\frac{1}{2}\right)}^{2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$