<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] *

*$B(0,0)$

*$P=6\,e^{i\alpha}=6\,(\cos\alpha+i\sin\alpha)\iff P_x=6\cos\alpha$ , and $P_y=6\sin\alpha$ , with $\alpha\in(0,{\pi\over3})$

*$A=A_x+iA_y$ , with $(A_x-P_x)^2+(A_y-P_y)^2=PA^2=10^2=100$ ; $\frac{A_y}{A_x}=\tan\frac\pi3=\sqrt3$

*$C(C_x,0)$ , with $(C_x-P_x)^2+P_y^2=PC^2=7^2=49$

*Heron's Formula : $S_{ABC}=\sqrt{s(s-a)(s-b)(s-c)}\ $ , with $\ s = \frac{a\ +\ b\ +\ c}2$ , where :

*$a=BC=C_x$

*$b^2=AC^2=(A_x-C_x)^2+A_y^2$

*$c^2=AB^2=A_x^2+A_y^2$

*Good Luck ! :-)

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