In $\triangle ABC$, we have $\angle C=3\angle A, a=27$ and $c=48$. What is $b$?

$\mathrm{(A)\ } 33 \qquad \mathrm{(B) \ }35 \qquad \mathrm{(C) \  } 37 \qquad \mathrm{(D) \  } 39 \qquad \mathrm{(E) \  }\text{not uniquely determined}$
From the Law of Sines, we have $\frac{\sin(A)}{a}=\frac{\sin(C)}{c}$, or $\frac{\sin(A)}{27}=\frac{\sin(3A)}{48}\implies 9\sin(3A)=16\sin(A)$.
We now need to find an identity relating $\sin(3A)$ and $\sin(A)$. We have 

$\sin(3A)=\sin(2A+A)=\sin(2A)\cos(A)+\cos(2A)\sin(A)$

$=2\sin(A)\cos^2(A)+\sin(A)\cos^2(A)-\sin^3(A)=3\sin(A)\cos^2(A)-\sin^3(A)$

$=3\sin(A)(1-\sin^2(A))-\sin^3(A)=3\sin(A)-4\sin^3(A)$.

Thus we have $9(3\sin(A)-4\sin^3(A))=27\sin(A)-36\sin^3(A)=16\sin(A)$
$\implies 36\sin^3(A)=11\sin(A)$.
Therefore, $\sin(A)=0, \frac{\sqrt{11}}{6},$ or $-\frac{\sqrt{11}}{6}$. Notice that we must have $0^\circ180^\circ$. We can therefore disregard $\sin(A)=0$ because then $A=0$ and also we can disregard $\sin(A)=-\frac{\sqrt{11}}{6}$ because then $A$ would be in the third or fourth quadrants, much greater than the desired range.

Therefore, $\sin(A)=\frac{\sqrt{11}}{6}$, and $\cos(A)=\sqrt{1-\left(\frac{\sqrt{11}}{6}\right)^2}=\frac{5}{6}$. Going back to the Law of Sines, we have $\frac{\sin(A)}{27}=\frac{\sin(B)}{b}=\frac{\sin(\pi-3A-A)}{b}=\frac{\sin(4A)}{b}$.

We now need to find $\sin(4A)$.
$\sin(4A)=\sin(2\cdot2A)=2\sin(2A)\cos(2A)$
$=2(2\sin(A)\cos(A))(\cos^2(A)-\sin^2(A))$
$=2\cdot2\cdot\frac{\sqrt{11}}{6}\cdot\frac{5}{6}\left(\left(\frac{5}{6}\right)^2-\left(\frac{\sqrt{11}}{6}\right)^2\right)=\frac{35\sqrt{11}}{162}$.

Therefore, $\frac{\frac{\sqrt{11}}{6}}{27}=\frac{\frac{35\sqrt{11}}{162}}{b}\implies b=\frac{6\cdot27\cdot35}{162}=35, \boxed{\text{B}}$.