Determing the voltage across a resistor
Let's call the bottom node "ground" and mark it as 0 V. The top node can be called voltage \$V_x\$, then. You know that the current into the node must be equal the current flowing back out (current doesn't accumulate in a node.) So the following must be true:

$$(1)~~~~~~~~~~~~30 \textrm{A} = \frac{V_x}{2.4\:\Omega} + \frac{V_x}{1.6\:\Omega+8\:\Omega} $$

In the above, you can see the two branch currents listed on the right side and the incoming 30 A current on the left side. They have to be equal.

The above equation solves into:

$$(2)~~~~~~~~~~~~~~~~~~~~~V_x=\frac{30 \textrm{A}}{\frac{1}{2.4\:\Omega} + \frac{1}{1.6\:\Omega+8\:\Omega}} $$

But you don't care about \$V_x\$. Just just want the expression that is right-most in the first equation listed above, namely the current through the branch with the two resistors. So you can plug one into the other to get:

$$I=\frac{\left(\frac{30 \textrm{A}}{\frac{1}{2.4\:\Omega} + \frac{1}{1.6\:\Omega+8\:\Omega}}\right)}{1.6\:\Omega+8\:\Omega}=\frac{30 \textrm{A}}{\frac{1.6\:\Omega+8\:\Omega}{2.4\:\Omega} + 1}=\frac{30 \textrm{A}}{\frac{1.6\:\Omega+8\:\Omega+2.4\:\Omega}{2.4\:\Omega}}=30 \textrm{A}\frac{2.4\:\Omega}{1.6\:\Omega+8\:\Omega+2.4\:\Omega}$$

That merely confirms what you've already been told. But it doesn't provide intuition.

In a voltage divider, the voltage across a series element is equal to the total voltage minus all the voltage drops across the other series elements.

Here, we use resistance and compute \$I_T=\frac{V_T}{\sum R_i}\$ for the divider, so,

$$\begin{align*}
(3)~~~~~~~~~V_T &amp;= \sum^n_1 R_i I_T = I_T\sum^n_1 R_i \\
 \\
V_T &amp;= I_T\left[\sum^n_1 R_i\right] - R_x\cdot I_T + V_x \\
 \\
V_x &amp;= V_T - I_T\left(\left[\sum^n_1 R_i\right] - R_x\right) \\
 \\
V_x &amp;= V_T - \frac{V_T}{\sum^n_1 R_i}\left(\left[\sum^n_1 R_i\right] - R_x\right) \\
 \\
V_x &amp;= V_T\cdot\left(1 - \frac{1}{\sum^n_1 R_i}\left(\left[\sum^n_1 R_i\right] - R_x\right)\right) \\
 \\
V_x &amp;= V_T\cdot\left(1 - \left(1 - \frac{R_x}{\sum^n_1 R_i}\right)\right) \\
 \\
V_x &amp;= V_T\cdot\frac{R_x}{\sum^n_1 R_i}
\end{align*}$$

In a current divider, the current through a parallel element is equal to the total current minus all the currents through other parallel elements.

Here, we use conductance and compute \$V_T=\frac{I_T}{\sum G_i}\$ for the divider (where \$G_i\$ is the conductance of each branch), so,

$$\begin{align*}
(4)~~~~~~~~~I_T &amp;= \sum^n_1 G_i V_T = V_T\sum^n_1 G_i \\
 \\
I_T &amp;= V_T\left[\sum^n_1 G_i\right] - G_x\cdot V_T + I_x \\
 \\
I_x &amp;= I_T - V_T\left(\left[\sum^n_1 G_i\right] - G_x\right) \\
 \\
I_x &amp;= I_T - \frac{I_T}{\sum^n_1 G_i}\left(\left[\sum^n_1 G_i\right] - G_x\right) \\
 \\
I_x &amp;= I_T\cdot\left(1 - \frac{1}{\sum^n_1 G_i}\left(\left[\sum^n_1 G_i\right] - G_x\right)\right) \\
 \\
I_x &amp;= I_T\cdot\left(1 - \left(1 - \frac{G_x}{\sum^n_1 G_i}\right)\right) \\
 \\
I_x &amp;= I_T\cdot\frac{G_x}{\sum^n_1 G_i}
\end{align*}$$

The symmetry should be obvious. In your case, you only have two conductances, \$G_1=\frac{1}{2.4\:\Omega}\$ and \$G_2=\frac{1}{1.6\:\Omega+8\:\Omega}\$. And in this case, \$G_x=G_2\$.

If you use those in the bottom equation of (4), you will get the right answer. However, it doesn't look like what the teacher wrote up. But if you plug things in, you can see why:

$$\begin{align*}
(5)~~~~~~~~~I_x &amp;= I_T\cdot\frac{\frac{1}{1.6\:\Omega+8\:\Omega}}{\frac{1}{2.4\:\Omega}+\frac{1}{1.6\:\Omega+8\:\Omega}} \\
 \\
&amp;= I_T\cdot\frac{\frac{1}{1.6\:\Omega+8\:\Omega}}{\frac{2.4\:\Omega+1.6\:\Omega+8\:\Omega}{2.4\:\Omega\cdot\left(1.6\:\Omega+8\:\Omega\right)}} \\
 \\
&amp;=I_T\cdot\frac{\frac{1}{1}}{\frac{2.4\:\Omega+1.6\:\Omega+8\:\Omega}{2.4\:\Omega}}=I_T\cdot\frac{2.4\:\Omega}{2.4\:\Omega+1.6\:\Omega+8\:\Omega}
\end{align*}$$

However, that is NOT a general approach. It just happens to work in this case. So your intuition perhaps is right, about being confused in class. The general solution looks like this, using resistances:

$$\begin{align*}
(6)~~~~~~~~~I_x &amp;= I_T\cdot\frac{G_x}{\sum^n_1 G_i} \\
  \\
&amp;= I_T\cdot\frac{\prod_{i=1,\\ i\ne x}^n R_i}{\sum^n_{i=1} \prod_{j=1,\\ j\ne i}^n R_j}
\end{align*}$$

Keep in mind that in the above, each \$R_i\$ is not each of all your resistors, helter-skelter. These resistances are the effective branch resistances. So in one of your branches, this value is a sum. Don't forget that, either.

Equation (6) is a lot uglier than you might have imagined from your teacher's solution. It's the correct one, though. So, if you want to commit something to memory, go with the conductances approach. As you can see, using resistances isn't right off the bat intuitive.

If you must use resistances in the current branching case like this, then you have to follow (6) above and DO NOT simply try and extend what your teacher did in some ignorant way. It won't work.



This means I don't agree an answer suggesting, "Your professor has given the formula for a current divider."

Wrong!

Normally, people do as some of the other answers suggest -- work out the series and parallel resistances to a single value (if possible) and then figure out the voltage. Then work through the details for the branch itself to get the current. That's pretty much how most would approach it. (Or use nodal analysis, if the circuit is more complex.)

Your teacher's formula isn't an extensible formula. It's just one that fell out in this case.