inscribed triangle

Since one of the side of the triangle is the diameter of the circle, it is a right triangle. Let $AB=2r$ and $\angle ABC=\theta$ , then $AC=2r \sin \theta$ and $BC=2r \cos \theta$ .So the area of the triangle is $\dfrac{1}{2}(2r\sin \theta)(2r \cos \theta)=2r^2(\sin \theta)(\cos \theta)$ . Using the double angle identity, $\sin~2x=2\sin~x\cos~x$ , we have

$A_T=r^2 \sin 2\theta$

Given in the problem that $\dfrac{A_C}{A_T}=2 \pi$ , so

$\dfrac{\pi r^2}{r^2 \sin~2\theta}=2 \pi$

$\dfrac{1}{2}=\sin 2\theta$

$2\theta=\sin^{-1}\left(\dfrac{1}{2}\right)$

$2 \theta = 30$

$\color{#D61F06}\boxed{\theta =15}$