Area of a Triangle Inside a Circle? I have a circle of radius $r$ and a triangle inside that circle. Specifically, if you have the triangle $\triangle ABC$ inside a circle with only the one side $AB$ and an angle $\angle \text{B}$ opposite to the other side $AC$ known along with circle radius $r$, could you determine its area?

 A: The angle at $C$ must be $\arcsin\frac{|AB|}{2r}$, namely half of the arc spanned by $AB$. This gives you all the angles in the triangle.
The law of sines then gives you the sides.
And then there are many ways to get the area, such as Heron's formula, or the sine of one angle times half the product of the adjacent sides.
A: Let $O$ be the center of the circle.
Let $\alpha=\angle ABO$ and $\beta=\angle OBC$
Then $\sin(\angle B)=\sin(\alpha)\cdot \cos(\beta)+\sin(\beta)\cdot \cos(\alpha)$. 
Of course $\cos(\alpha)=\frac{|AB|}{2r}$ and $\sin(\alpha)=\sqrt{1-\cos^2(\alpha)}$.
Therefore if $x=\cos(\beta)$ we get
$\sin(\angle B)=\sqrt{1-(\frac{|AB|}{2r})^2}\cdot x + \frac{|AB|}{2r}\sqrt{1-x^2}$.
So we solve the equation for $x$ to get $\cos(\beta)$ (don't get scared, if you take the first summand to the left side and square it is just a quadratic polynomial in $x$).
And once we find $\cos(\beta)$ we have $|BC|=2r\cdot \cos(\beta)$
Finally, the area is just $\sin(\angle B)\cdot\frac{|BC||BA|}{2}$
