Use the image to answer the question.

Question

A circle has center point upper F. The circumference point upper L is at the left of the circle and the circumference point upper M is at the right of the circle. Point upper K is above the circle in line with upper F.

Identify and use the relationship between the radius and a line tangent to the circle, where the radius intersects the circle, to solve the following. If KM¯¯¯¯¯¯¯¯¯¯
and KL¯¯¯¯¯¯¯¯
are tangent to the circle at L
and M
, ∠LFK=23 degrees
, and KM¯¯¯¯¯¯¯¯¯¯=47 feet
, how long is LF¯¯¯¯¯¯¯
to the nearest foot?

(1 point)
Responses

51 ft.
51 ft.

111 ft.
111 ft.

20 ft.
20 ft.

120 ft.

Answer 1

To find the length of the radius \( LF \) of the circle, we'll use the information provided about the tangents and the angle.

1. Understand the relationship: The tangent lines \( KL \) and \( KM \) are perpendicular to the radius \( LF \) at points \( L \) and \( M \) respectively. This means we have two right triangles, \( LFK \) and \( LFM \).

2. Identify the triangle: In triangle \( LFK \), we have:

   • \( \angle LFK = 90^\circ - \angle LFK = 90^\circ - 23^\circ = 67^\circ \)
   • We know \( KM = 47 \) feet (the length of the tangent segment from point \( K \) to point \( M \)).

3. Using the tangent-secant theorem and trigonometry: The length of the tangent \( KL \) can be identified with the opposite side of angle \( LFK \) in triangle \( LFK \): \[ KL = LF \cdot \tan(\angle LFK) \] However, since we want to solve for \( LF \), with \( KL = 47 \) feet, we can express our tangent relationship: \[ 47 = LF \cdot \tan(23^\circ) \]

4. Rearranging for \( LF \): \[ LF = \frac{47}{\tan(23^\circ)} \]

5. Calculating the values: Using a calculator, find \( \tan(23^\circ) \): \[ \tan(23^\circ) \approx 0.4245 \] Therefore: \[ LF = \frac{47}{0.4245} \approx 110.72 \]

6. Rounding to the nearest foot: Rounding \( 110.72 \) gives \( 111 \).

Thus, the length of \( LF \) to the nearest foot is:

111 ft.