13.56 The sides of a triangle are three consecutive integers, and the largest angle is twice the smallest angle. Then the cosine of the smallest angle is
(A) $\frac{3}{4}$.
(B) $\frac{7}{10}$.
(C) $\frac{2}{3}$.
(D) $\frac{9}{14}$.
(E) None of these values.
(33rd American High School Mathematics Examination, 1982)
[Solution] As shown in the figure, let's assume that the smallest angle opposite the shortest side $n$ is $\theta$. By the Law of Sines,
we have $\frac{\sin \theta}{n}=\frac{\sin 2 \theta}{n+2}$, which means $\frac{\sin \theta}{n}=\frac{2 \sin \theta \cos \theta}{n+2}$,
so $\cos \theta=\frac{n+2}{2 n}$.
By the Law of Cosines, we have

Therefore,
$$
\begin{aligned}
\cos \theta & =\frac{(n+1)^{2}+(n+2)^{2}-n^{2}}{2(n+1)(n+2)}, \\
\frac{n+2}{2 n} & =\frac{(n+1)^{2}+(n+2)^{2}-n^{2}}{2(n+1)(n+2)} \\
& =\frac{(n+1)(n+5)}{2(n+1)(n+2)}=\frac{n+5}{2(n+2)} .
\end{aligned}
$$

Solving for $n$ gives $n=4$, so $\cos \theta=\frac{4+2}{4 \cdot 2}=\frac{3}{4}$.
Therefore, the answer is $(A)$.