<s>[INST] 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 
 [/INST] let $B$ is on the $O$ of polar system, $P(6,\alpha)$ 
so $C$ is on circle $p^2+6^2-2*6*cos(\beta- \alpha)=7^2$ and $\beta=0$
$A$ is on circle $p^2+6^2-2*6*cos(\beta- \alpha)=10^2$ and $\beta=\dfrac{\pi}{3}$
$a=BC=6cos\alpha+\sqrt{(6cos\alpha)^2+13}, c=BA=6cos(\dfrac{\pi}{3}-\alpha)+\sqrt{(6cos(\dfrac{\pi}{3}-\alpha))^2+64}$
$ac=f(\alpha),\alpha=x, f'(x)=(\sqrt{36 cos^2(x)+13}+6 cos(x)) \left(6 cos(x+\dfrac{\pi}{6})+\dfrac{(36 sin(x+\dfrac{\pi}{6}) cos(x+\dfrac{\pi}{6})}{\sqrt{36 sin^2(x+\dfrac{\pi}{6})+64)}}\right)$$+(\sqrt{36 sin^2(x+\dfrac{\pi}{6})+64)+6 sin(x+\dfrac{\pi}{6}}) \left(-6 sin(x)-\dfrac{(36 sin(x) cos(x))}{\sqrt{36 cos^2(x)+13)}}\right)=0$
it is only numeric method can solve the solution is $x=.422064 \implies f_{max}=171.138$
BTW, this triangle also have max area without fixed $\angle B$ but it is more difficult as there is two varies.  
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