What is the greatest possible area of a triangular region with one vertex at the center of a circle of radius 1 and the other two vertices on the circle?
A bad first step is to put the center at the origin, one point at (1,0) , and one point at (sin x, cos x).
A start is the area of a triangle with included angle expression,
$$ {a \times b \times \sin \theta} \over {2}$$
Assuming $\theta$ in radians. If theta is $\pi/2$ then we have a right triangle. Let a=b=1.
Area expression is $$A=(\sin \theta) / 2$$ This is maximum for $\theta = \pi/2$.
Answer is maximum area for a right triangle.
Best Answer
The area of a triangle with side lengths $a, b, c$ and angles $A, B, C$ can be expressed as $$\left|\triangle ABC\right| = \frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A = \frac{1}{2}ca \sin B.$$ Thus the area for a fixed $a = b = 1$ and variable central angle $C$ is maximized when $\sin C$ attains its maximum; i.e., $\angle C = \pi/2$ and $\sin C = 1$. Thus the triangle is right.