1. trig with no calc

in triangle ABC, angle C = $\displaystyle (\frac{\pi}{6})$ and the side opposite angle c, AB = 2

I know $\displaystyle (\frac{\pi}{6})$ = 30 degrees, and can solve this easily with a calculator using tan but how do you solve for that side without one?

Thanks

2. You should also know that $\displaystyle \tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \: \text{ or } \: \frac{\sqrt{3}}{3}$

So: $\displaystyle \frac{1}{\sqrt{3}} = \frac{\text{AB}}{\text{AC}}$

Assuming this is a right triangle ...

3. Originally Posted by o_O
You should also know that $\displaystyle \tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \: \text{ or } \: \frac{\sqrt{3}}{3}$

So: $\displaystyle \frac{1}{\sqrt{3}} = \frac{\text{AB}}{\text{AC}}$

Assuming this is a right triangle ...
could you explain

well, $\displaystyle tan = \frac{sin}{{cos}}$ = $\displaystyle \frac{\frac{1}{2}}{\frac{2}{\sqrt{3}}} = \frac{1}{2}*\frac{2}{\sqrt{3}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$

ha

thanks for the push

4. Do you know your unit circle? i.e. the sin, cos, tan values of certain angles and $\displaystyle \frac{\pi}{6}$ should definitely be one of them.

So, we have:
$\displaystyle \begin{array}{rcl}\displaystyle\tan \left(\frac{\pi}{6}\right) & = & \displaystyle\frac{\text{AB}}{\text{AC}} \\& & \\ \displaystyle \frac{1}{\sqrt{3}} & = & \displaystyle\frac{2}{\text{AC}} \end{array}$