$7 \cdot 43$ In $\triangle A B C$, $\angle C=3 \angle A$, $a$ $=27$, and $c=48$. Then $b$ equals
(A) 33 .
(B) 35 .
(C) 37 .
(D) 39 .
(E) The value of $b$ is not unique.
[Solution 1] Let $\angle A C D = \angle A$, then $\angle B C D = \angle B D C = 2 \angle A$.
$\therefore B D = B C = 27$, and $C D = A D = 48 - 27 = 21$.
Construct the angle bisector $D E$ of $\angle C D B$, then in $\triangle B D C$, the angle bisector of $\angle C D B$
$$
\begin{aligned}
D E & =\sqrt{21 \cdot 27\left[1-\left(\frac{27}{21+27}\right)^{2}\right]} \\
& =\sqrt{\frac{21 \cdot 27 \cdot\left(16^{2}-9^{2}\right)}{16^{2}}} \\
& =\frac{5 \cdot 7 \cdot 9}{16} .
\end{aligned}
$$

Also, $\angle E D C = \angle A C D = \angle A$,
$\therefore \quad D E \parallel A C$.
$\therefore \quad \frac{A C}{D E}=\frac{A B}{B D}$,
i.e., $A C=\frac{D E \cdot A B}{B D}=\frac{5 \cdot 7 \cdot 9 \cdot 48}{16 \cdot 27}=35$.
Therefore, the answer is $(B)$.
[Solution 2] By the Law of Sines and the given conditions,

and
then
then
i.e.,
solving gives
solving gives
By $C D + A D > A C$, we know 45 is not valid and should be discarded. Thus, $b=35$. Therefore, the answer is $(B)$.
[Solution 3] Construct $\angle A C D = \angle A$, then $\angle D C B = 2 \angle A, \quad \angle C D B = 2 \angle A$, $\therefore \quad B D = B C = 27$,
and $C D = A D = 21$.
Then $\cos \angle C D B = \frac{21^{2} + 27^{2} - 27^{2}}{2 \cdot 21 \cdot 27} = \frac{21}{54} = \frac{7}{18}$,
$\therefore \cos \angle A D C = -\frac{7}{18}$, and by the Law of Cosines,
$$
\begin{array}{l}
b^{2} = A D^{2} + C D^{2} - 2 A D \cdot C D \cdot \cos \angle A D C \\
= 21^{2} + 21^{2} + 7^{3} = 7^{2}\left(3^{2} + 3^{2} + 7\right) \\
= 7^{2} \cdot 5^{2}, \\
\therefore \quad b = 35. \\
\end{array}
$$

Therefore, the answer is $(B)$.
[Solution 4] As shown in the figure, with $B$ as the center and $B C$ as the radius, construct a circle, and connect $G C$.
It is easy to prove
$$
\begin{aligned}
& \angle B G C = \angle B C G = \frac{1}{2}\left(180^{\circ} - \angle B\right) \\
= & \frac{1}{2}\left[180^{\circ} - \left(180^{\circ} - 4 \angle A\right)\right] \\
= & 2 \angle A .
\end{aligned}
$$
$\therefore \angle G C A = \angle A$.
Again, with $B$ as the center and $B A$ as the radius, construct a circle, extend $A C$ and $B C$ to intersect the larger circle at points $D, E, F$, and connect $B D$. We can get $\angle D = \angle A = \angle G C A$, then $G C \parallel B D$.
$\therefore \frac{A C}{A D} = \frac{A G}{A B}$, i.e., $\frac{b}{b + C D} = \frac{21}{48}$,
solving gives $C D = \frac{9}{7} b$.
Also, $E C = 21, C F = 75$. By the intersecting chords theorem,
$A C \cdot C D = E C \cdot C F$, i.e., $C D = \frac{E C \cdot C F}{A C} = \frac{21 \cdot 75}{b}$.
$\therefore \frac{9}{7} b = \frac{21 \cdot 75}{b}$, giving $b^{2} = 7^{2} \cdot 5^{2}$, i.e., $b = 35$.
Therefore, the answer is $(B)$.