How will you do it?

Call the vertices of our triangle $A$ , $B$ , and $C$ and call the center of the circle, $O$ . Construct radii $\overline{OA}$ , $\overline{OB}$ , and $\overline{OC}$ . By the inscribed angle theorem, the measure of each central angle is double that of its corresponding inscribed angle. Thus $\begin{aligned} m\angle BOC &= 83^\circ \\ m\angle AOC &= 141^\circ \\ m\angle AOB &= 136^\circ \end{aligned}$ The area of any triangle can be found from two sides and the included angle using the formula $A=\frac{1}{2}s_1s_2\sin \theta$ , where $\theta$ is the angle included between sides $s_1$ and $s_2$ (for those wishing to prove this, start by constructing an altitude from angle $C$ ). Thus $\begin{aligned} \mathrm{Area}(\triangle BOC) &= \frac{1}{2}\cdot 4^2 \cdot \sin 83^\circ = 8\sin 83^\circ \\ \mathrm{Area}(\triangle AOC) &= \frac{1}{2}\cdot 4^2 \cdot \sin 141^\circ = 8\sin 141^\circ \\ \mathrm{Area}(\triangle AOB) &= \frac{1}{2}\cdot 4^2 \cdot \sin 136^\circ = 8\sin 136^\circ \end{aligned}$ Finally, $\mathrm{Area}(\triangle ABC) = \mathrm{Area}(\triangle BOC) + \mathrm{Area}(\triangle AOC) + \mathrm{Area}(\triangle AOB) = 8\sin 83^\circ + 8\sin 141^\circ + 8\sin 136^\circ \approx \boxed{18.532}$

Use formula area = 2 (r^2) Sin(A) Sin(B) Sin(C)