$7 \cdot 54$ In $\triangle A B C$, if $A B=2 B C, \angle B=2 \angle A$, then $\triangle A B C$ is
(A) an acute triangle.
(B) a right triangle.
(C) an obtuse triangle.
(D) cannot be determined.
(China Sichuan Province Junior High School Mathematics Competition, 1992)
［Solution］From the given and the Law of Sines, we have
$$
\frac{1}{\sin A}=\frac{2}{\sin \left(180^{\circ}-3 A\right)},
$$

which means
$$
\begin{array}{l}
2 \sin A=\sin 3 A, \\
2 \sin A=3 \sin A-4 \sin ^{3} A, \\
\sin A\left(1-4 \sin ^{2} A\right)=0,
\end{array}
$$
$$
\sin A=0 \text { or } 1-4 \sin ^{2} A=0 \text {. }
$$
Solving, we get $\angle A=0^{\circ}$ (discard) or $\angle A=30^{\circ}$.
Then $\angle B=60^{\circ}$, so $\triangle A B C$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ right triangle. Therefore, the answer is (B).