Maximal area of triangle if angle and opposite side length is known Two lines $l_1$ and $l_2$ intersects at point $A$ such that the angle they intersect is $\alpha$. A line segment has endpoints $B$ and $C$ in the lines $l_1$ and $l_2$, respectively, and $|BC|=l$. What is the maximal area of $ABC$ in terms of $\alpha$ and $l$ if those two variables are fixed?
 A: There are several ways to parametrize this problem and therefore several expressions for the target function. I will express it in terms of a variable angle.

Denote one of the variable sides, say $AC$, $x$ and $\angle ACB=\beta$. The area of $\triangle ABC$ is half the product of two sides times the sine of the enclosed angle:
$$S=\frac{1}{2}xl\sin\beta$$
From the sine rule we deduce the following relation:
$$\frac{l}{\sin\alpha}=\frac{x}{\sin\gamma}$$
Substituting in the expression for $S$:
$$S=\frac{l^{2}}{2\sin\alpha}\sin\beta\sin\gamma$$
So the problem reduces to the following:
$$f=\sin\beta\sin\gamma\to\max$$
subject to the constraint:
$$\beta+\gamma=\pi-\alpha$$
Simplify using complimentary formula:
$$f\left(\beta\right)=\sin\beta\sin\left[\pi-\left(\alpha+\beta\right)\right]=\sin\beta\sin\left(\alpha+\beta\right)$$
$$f'\left(\beta\right)=\sin\left(\alpha+\beta\right)\cos\beta+\sin\beta\cos\left(\alpha+\beta\right)=\sin\left(\alpha+2\beta\right)$$
Find stationary points:
$$f'\left(\beta\right)=0$$
$$\sin\left(\alpha+2\beta\right)=0$$
From the graph of sine and the consideration that $0\le\alpha,\beta\le \pi$ we obtain:
$$\alpha+2\beta=\pi$$
$$\beta=\frac{\pi-\alpha}{2}$$
$$\gamma=\pi-\alpha-\frac{\pi-\alpha}{2}=\frac{\pi-\alpha}{2}$$
Hence the triangle is isosceles. Now the area can be easily computed.
A: Consider any such triangle $\triangle A'BC$, not necessary with maximal area. Then any other such triangle must have $A$ lying on the circumcircle of $\triangle A'BC$, and on the arc $BA'C$. Now, consider the line through $A$ parallel to $BC$. For the area to be maximal, this line must be as far away from $BC$ as possible. This obviously happens when it is tangent to the circle. In this case, The diameter through $A$ is perpendicular to $BC$, and hence is the perpendicular bisector of $BC$. So $AB = AC$.
