# A triangle has sides A, B, and C. The angle between sides A and B is (pi)/6 and the angle between sides B and C is pi/6. If side B has a length of 2, what is the area of the triangle?

Apr 23, 2018

No! Stop doing that! A triangle doesn't have sides A, B and C! Triangle ABC has vertices A, B and C and corresponding opposing sides $a , b$ and $c .$ It's much easier if we do it like this every time.

This problem should read: What is the area of a triangle ABC labeled in the usual way with $A = C = \frac{\pi}{6}$ and $b = 2$?

I really don't like trig as taught in school because every problem is 30,60,90 or 45,45,90. It's like we have a whole subject that only works for two triangles.

Boy I'm grumpy today. I'll stop griping and just do the problem.

We have $A = C = \frac{\pi}{6} = {30}^{\circ}$. That's an isosceles triangle. So half of it will be a right triangle, with side $\frac{b}{2}$, side (and altitude of the original triangle) $h$, which satisfies

$h = \frac{b}{2} \tan A$

Call the area $m a t h c a l \left\{A\right\}$.

$m a t h c a l \left\{A\right\} = \frac{1}{2} b h = {b}^{2} / \left\{4 \tan A\right\} = \frac{{2}^{2}}{4 \tan 30} = \setminus \sqrt{3}$