How find this maximum $S_{\Delta ABC}$ in $\Delta ABC$,and $\angle ABC=60$,such that $PA=10,PB=6,PC=7$,
find the maximum $S_{\Delta ABC}$.

My try:let $AB=c,BC=a,AC=b$, then
$$b^2=a^2+c^2-2ac\cos{\angle ABC}=a^2+c^2-2ac$$
then
$$S_{ABC}=\dfrac{1}{2}ac\sin{60}=\dfrac{\sqrt{3}}{4}ac$$
Then I can't 

IN $\bigtriangleup$ABC we take AB=c;AC=b &BC=a


*

*NOW $a^2$ = $6^2$+$7^2$ -2.6.7.cos$\angle$BPC

*NOW cos$\theta$ $\geq$ -1$\longrightarrow$   -cos$\theta$$\leq$1. from this we get $a^2$$\leq$13

*$c^2$= $10^2$+$6^2$ -2.10.6.cos$\angle$APB.similarly from this we get [applying the inequality] c$\leq$16

*now the area of $\bigtriangleup$ABC=$\frac{1}{2}$acsin60=$\sqrt[2]{3}$.$\frac{1}{4}$.ac$\leq$$\sqrt[2]{3}$.$\frac{1}{4}$.16. 13=52$\sqrt[2]{3}$
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