# maximizing the area of a triangle with given angle and opposite side

Let $\Delta ABC$ have given angle $\theta$ at vertex A and opposite side BC of given length $a$.

I would like to find the maximum area of such a triangle in terms of $\theta$ and $a$.

I think the triangle of maximum area is isosceles, but how can I show this (if this is correct)?

(I realize that this can also be worked using calculus, but I am looking for a more elementary solution.)

• Hypothesis: Given $\triangle ABC$ as in your answer, let $X$ be the circle passing through $A$, $B$, $C$. The points $B,C$ cut $X$ into two circular arcs; let $\alpha$ be the one containing $A$.
• Conclusion: for every $A' \in \alpha$ the angles of $\triangle ABC$ and $\triangle A'BC$ at $A,A'$ are equal.
Hence, your problem is solved by choosing $A' \in \alpha$ so as to maximize the height of the triangle $A'BC$ with base $\overline{BC}$. That choice of $A'$ is unique, namely the intersection of $\alpha$ with the perpendicular bisector of $\overline{BC}$, and that does indeed give an isosceles triangle $A'BC$.
• Thanks for your answer. (The way I set up the notation, would I want to use the arcs cut out by the points B and C, and let $\alpha$ be the arc containing A?) – user84413 Dec 23 '15 at 18:01