Question

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?

Answer 1

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=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=pi/6=30^circ`. That's an isosceles triangle. So half of it will be a right triangle, with side `b/2`, side (and altitude of the original triangle) `h`, which satisfies

`h = b/2 tanA`

Call the area `mathcal{A}`.

`mathcal{A} = 1/2 b h = b^2/{4 tan A} = {2^2}/{4 tan 30} =\sqrt{3}`