7.39 In $\triangle A B C$, if $\angle A=60^{\circ}, A C=16$, and the area of the triangle is $220 \sqrt{3}$, then the length of $B C$ is
(A) $\sqrt{2400}$.
(B) 25 .
(C) 51 .
(D) $49 \sqrt{3}$.
(E) 49 .
(China Beijing Junior High School Mathematics Competition, 1982)
[Solution] From $S_{\triangle A B C}=\frac{1}{2} A C \cdot A B \sin A$ we get
$$
\begin{array}{cc} 
& \frac{1}{2} \cdot 16 \cdot A B \sin 60^{\circ}=220 \sqrt{3}, \\
\therefore \quad & A B=55 .
\end{array}
$$

From the cosine rule we get
$$
\begin{array}{l} 
B C^{2}= A C^{2}+A B^{2}-2 A C \cdot A B \cos 60^{\circ} \\
= 16^{2}+55^{2}-16 \cdot 55=2401 . \\
\therefore \quad B C=\sqrt{2401}=49 .
\end{array}
$$

Therefore, the answer is $(E)$.