The area and angle $θ$ of a triangle are given, when the side opposites to the given angle is minimum, then the length of remaining two sides are
I used sine rule and equation for area, but not getting any relation. The answer must be in terms of square root of area multiplied with a constant divided by sine of the given angle.
The problem is re-written as follows:
Given $\theta, A$ and $ab\sin(\theta)=2A$, find the value of $a,b$ that minimizes $a^2+b^2-2ab\cos(\theta)$.
We note that $b={2A\over\sin(\theta)a}$
Therefore $a^2+b^2-2ab\cos(\theta)=a^2+{4A^2\over\sin(\theta)^2a^2}-{4A\cos(\theta)\over \sin(\theta)}$.
Note the last term is constant and the minimum occurs at $a^2={4A^2\over\sin(\theta)^2a^2}$ by $AM-GM$. By symmetry we have similar argument to show $b^2={4A^2\over\sin(\theta)^2b^2}$ as well and finally $a=b=\sqrt{2A\over \sin(\theta)}$