Answer:
To solve this problem, we can use the properties of tangents and secants intersecting circles.
Given:
- \( AM = 6 \) cm
- \( m(\angle LB) = 30^\circ \)
We want to find \( MB \).
Since \( AB \) is tangent to the circle at point \( B \), and \( AM \) intersects the circle at point \( M \), we can use the tangent-secant theorem, which states that the length of the secant segment (in this case, \( AM \)) multiplied by its external segment (in this case, \( MB \)) is equal to the square of the length of the tangent segment (in this case, \( AB \)).
So, we have:
\[ AB^2 = AM \times MB \]
\[ AB^2 = 6 \times MB \]
To find \( AB \), we can use trigonometry. Since \( \angle LBM = 90^\circ \) (angle between tangent and radius is always 90 degrees), we can use trigonometric ratios:
\[ \tan(30^\circ) = \frac{MB}{AB} \]
\[ \frac{1}{\sqrt{3}} = \frac{MB}{AB} \]
\[ AB = \frac{MB}{\frac{1}{\sqrt{3}}} = MB \sqrt{3} \]
Now, we can substitute \( AB = MB \sqrt{3} \) into our equation:
\[ (MB \sqrt{3})^2 = 6 \times MB \]
\[ 3MB^2 = 6MB \]
\[ 3MB^2 - 6MB = 0 \]
\[ 3MB(MB - 2) = 0 \]
This equation has two solutions: \( MB = 0 \) (which is not applicable in this context) and \( MB = 2 \). However, since we're dealing with a length, we discard the solution \( MB = 0 \).
Therefore, \( MB = 2 \) cm.
So, the correct answer is (b) \( 6\sqrt{3} \) cm.
