Let $a \in \left( \frac { \pi } { 6 } , \frac { \pi } { 4 } \right)$. Given that
$$\begin{aligned} & x = \tan a \\ & y = \tan ( 2 a ) \\ & z = \tan ( 3 a ) \end{aligned}$$
Which of the following is the correct ordering of these numbers?
A) $x < y < z$
B) $x < z < y$
C) $y < x < z$
D) $z < x < y$
E) $z < y < x$
Let $a \in \left( \frac { \pi } { 6 } , \frac { \pi } { 4 } \right)$. Given that

$$\begin{aligned}
& x = \tan a \\
& y = \tan ( 2 a ) \\
& z = \tan ( 3 a )
\end{aligned}$$

Which of the following is the correct ordering of these numbers?\\
A) $x < y < z$\\
B) $x < z < y$\\
C) $y < x < z$\\
D) $z < x < y$\\
E) $z < y < x$