Right triangles
Initial explanation
We know that the tangent equation if given by:

In this case, with respect to 30º, the opposite side is AC and the adjacent is BC:
Tangent equation in this case
Then, in this case, we have

since tan(30º) = 0.577 and AC = 6, then:

Finding BC
Now, we can solve the equation for BC "leaving it alone".
Step 1- taking BC to the left side, we have:

Step 2- taking 0.577 to the right side:

Then,

Answer: BC = 10.4