1. Let $\triangle A B C$ be a right triangle with right angle at $B$. Let the points $D, E$, and $F$ be on $A B, B C$, and $C A$, respectively, such that $\triangle D E F$ is an equilateral triangle and $E C=F C$. If $D B=5 \sqrt{3}, B E=3$, and $\sin \angle A C B=4 \sqrt{3} / 7$, find the perimeter of $\triangle A D F$.
Answer: $35 \sqrt{3}+63+2 \sqrt{21}$
Solution: By Pythagorean Theorem, $D E=\sqrt{(5 \sqrt{3})^{2}+3^{2}}=\sqrt{84}=E F$. From $\sin \angle A C B$ $=4 \sqrt{3} / 7$, we have $\cos \angle A C B=1 / 7$. Let $E C=F C=x$, then by Cosine Law on side $E F$ of $\triangle E C F$, we have $E F^{2}=x^{2}+x^{2}-2 x^{2} \cos \angle A C B$. Solving for $x^{2}$, we have $x^{2}=84 /[2(1-1 / 7)]=49$. Hence, $E C=F C=x=7$ and $B C=B E+E C=10$. Since $\cos \angle A C B=1 / 7$, then $A C=70$, which means $A F=63$. Since $\sin \angle A C B=4 \sqrt{3} / 7$, then $A B=40 \sqrt{3}$, which means $A D=35 \sqrt{3}$. Thus, the perimeter of $\triangle A D F$ is $35 \sqrt{3}+63+2 \sqrt{21}$.