The lengths of the sides of a triangle are consecutive integers, and the largest angle is twice the smallest angle. 
The cosine of the smallest angle is 
$\textbf{(A)}\ \frac{3}{4}\qquad \textbf{(B)}\ \frac{7}{10}\qquad \textbf{(C)}\ \frac{2}{3}\qquad \textbf{(D)}\ \frac{9}{14}\qquad \textbf{(E)}\ \text{none of these}$
In $\triangle ABC,$ let $a=n,b=n+1,c=n+2,$ and $\angle A=\theta$ for some positive integer $n.$ We are given that $\angle C=2\theta,$ and we need $\cos\theta.$
We apply the Law of Cosines to solve for $\cos\theta:$ \[\cos\theta=\frac{b^2+c^2-a^2}{2bc}=\frac{n+5}{2(n+2)}.\]
We apply the Law of Sines to $\angle A$ and $\angle C:$ \[\frac{\sin\theta}{a}=\frac{\sin(2\theta)}{c}.\]
By the Double-Angle Formula $\sin(2\theta)=2\sin\theta\cos\theta,$ we simplify and rearrange to solve for $\cos\theta:$ \[\cos\theta=\frac{c}{2a}=\frac{n+2}{2n}.\]
We equate the expressions for $\cos\theta:$ \[\frac{n+5}{2(n+2)}=\frac{n+2}{2n},\]
from which $n=4.$ By substitution, the answer is $\cos\theta=\boxed{\textbf{(A)}\ \frac{3}{4}}.$
~MRENTHUSIASM