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 
 A: 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.  
A: *

*$B(0,0)$

*$P=6\,e^{i\alpha}=6\,(\cos\alpha+i\sin\alpha)\iff P_x=6\cos\alpha$ , 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 ! :-)

A: 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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