The point \((12, 3\sqrt{2})\) cannot be on the unit circle because the unit circle is defined by the equation \(x^2 + y^2 = 1\), and the coordinates of the point must satisfy this equation.
Let's check:
\[ 12^2 + (3\sqrt{2})^2 = 144 + 18 = 162 \neq 1. \]
This means we need to convert the point into a position relative to the unit circle. To find the corresponding sine of the angle, we should normalize the point \((12, 3\sqrt{2})\).
First, calculate the radius (the distance from the origin to the point):
\[ r = \sqrt{12^2 + (3\sqrt{2})^2} = \sqrt{144 + 18} = \sqrt{162} = 9\sqrt{2}. \]
Now, we can find the coordinates on the unit circle corresponding to this point by dividing each coordinate by the radius \(r\):
\[ \text{Normalized } x = \frac{12}{9\sqrt{2}} = \frac{4}{3\sqrt{2}} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}, \] \[ \text{Normalized } y = \frac{3\sqrt{2}}{9\sqrt{2}} = \frac{1}{3}. \]
Now we have the coordinates on the unit circle:
\((\frac{2\sqrt{2}}{3}, \frac{1}{3})\).
The sine of the angle is the \(y\)-coordinate of this normalized point. Therefore, the sine of the angle is:
\[ \sin(\theta) = \frac{1}{3}. \]
So, the answer is:
\(\frac{1}{3}\).
However, none of the given options contain \(\frac{1}{3}\). If this is an issue with the options, please confirm if the original point is indeed \((12, 3\sqrt{2})\) and check the correctness of the options as well.