Answer
See figure and explanatioins.
Work Step by Step
Let $AB$ be the $x$-axis and $C$ be on the unit circle with coordinates $C(x,y)$ in the first quadrant (including axis) as shown in the figure. We can express the area of the triangle as $A=\frac{1}{2}(2)(y)=y$. As $0\leq y\leq 1$, the maximum of $A$ can be obtained when $y=1$. This means that the triangle is isosceles with $C$ at $(0,1)$ and $AC=BC=\sqrt 2$.
