Area of a Triangle in a Circle (Given some angles)

Question

I inscribe a triangle in a circle of radius 1. WLOG, we can assume that the first point (A) of the triangle is $(1,0)$ or $\cos(0), \sin(0)$. Let the second point (B), be at $\cos(a), \sin(a)$, and the final point C be at $\cos(a+b), \sin(a+b)$. What's the area of the triangle?

I managed to figure out that $AB = 2\sin(a/2)$, $AC = 2\sin(b/2)$, $BC = 2\sin((a+b)/2)$.

Can I use this in a better way than Heron's formula to find out the area of the triangle, or is there just a better way to use $a$ and $b$ to find the area of the triangle? I really don't want to use Heron's formula, that seems incredibly ugly.

Answer 1

The signed area of the triangle with one vertex at $O$ and the other vertices on the circumference of a unit circle is $\frac12\sin\angle O$.

Your triangle is $$\triangle AOB + \triangle BOC-\triangle AOC\ ,\tag{$*$}$$ and its area is therefore $$\frac12(\sin a+\sin b-\sin(a+b))\ .$$

This is clear from a diagram if $a$ and $b$ are both acute angles. If $\pi<a+b<2\pi$ then the above gives a negative area for $\triangle AOC$, and the final area formula for $\triangle ABC$ remains correct.

Answer 2

David’s solution is slicker, but here’s a way that doesn’t require very much calculation: $$\begin{align} A &= \frac12 \begin{vmatrix}1&0&1 \\ \cos a & \sin a & 1 \\ \cos(a+b) & \sin(a+b) & 1 \end{vmatrix} \\ &= \frac12\left(\cos a \sin(a+b)-\sin a\cos(a+b) - \sin(a+b) + \sin a \right) \\ &=\frac12\left(\sin a + \sin b - \sin(a+b)\right). \end{align}$$