Triangle Area Sum 2

Through this solution, I will refer to law of cosines $a^2=b^2+c^2-2bc\cos(A)$ as LOC and area of a triangle in terms of sine $\frac{1}{2}ab\sin{C}$ as ATS. Also, lower case letters represent sides while uppercase letters represent angles.

Begin with the simple case of $a=11,~b=12~C=60$ .

$\text{ATS}\Longrightarrow \frac{1}{2}(11)(12)\left(\frac{\sqrt{3}}{2}\right)=33\sqrt{3}\approx 57.157$

Now if $a=11,~b=12,~A=60$

$\begin{aligned} \text{LOC}\longrightarrow a^2&=b^2+c^2-2bc\cos(A)\\ 121&=144+c^2-12c\\ 0&=c^2-12c+23\\ c&=6\pm \sqrt{13}\\ \text{ATS}& \Longrightarrow \frac{1}{2}(6\pm \sqrt{13})(12)\left(\frac{\sqrt{3}}{2}\right)\\ &=18\sqrt3\pm3\sqrt{39}\\ &\approx 49.9,~12.4 \end{aligned}$

Now if $a=11,~b=12,~B=60$

$\begin{aligned} \text{LOC}\longrightarrow b^2&=a^2+c^2-2ac\cos(B)\\ 144&=121+c^2-11c\\ 0&=c^2-11c-23\\ c&=\dfrac{11\pm\sqrt{213}}{2}\\ &\text{c can't be negative, so ignore the - case}\\ \text{ATS}& \Longrightarrow \frac{1}{2}\left(\frac{11+\sqrt{213}}{2}\right)(11)\left(\frac{\sqrt{3}}{2}\right)\\ &=\text{something ugly we don't care about}\\ &\approx 60.96\\ \end{aligned}$

We have areas $60.96>57.2>49.9>12.4$

Thus summing the three smallest we get $(33\sqrt{3})+(18\sqrt3+3\sqrt{39})+(18\sqrt3-3\sqrt{39})=69\sqrt3$

Therefore $a\sqrt{b}$

$a+b=69+3=72$