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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

120 ft.
120 ft.

20 ft.
20 ft.

111 ft.
111 ft.

51 ft.
51 ft.

Answer 1

To solve the problem, we need to analyze the right triangle formed by the radius \( LF \), the line \( KL \) (the tangent to the circle at point \( L \)), and the line from point \( F \) to point \( K \).

1. Understanding the setup:

   • The radius \( LF \) is perpendicular to the tangent line \( KL \) because a radius at the point of tangency is always perpendicular to the tangent line.
   • Therefore, triangle \( LFK \) is a right triangle with:
     • \( LF \) as one leg,
     • \( KL \) (which is equal to \( KM \)) as the other leg (the length of the tangent),
     • \( FK \) as the hypotenuse.

2. Given values in the problem:

   • \( \angle LFK = 23^\circ \)
   • \( KL = KM = 47 \) feet

3. Using trigonometry: In triangle \( LFK \):

   • \( \tan(23^\circ) = \frac{LF}{KL} \)
   • Rearranging gives us \( LF = KL \cdot \tan(23^\circ) \)

4. Calculating \( LF \):

   • Substitute \( KL \): \[ LF = 47 \cdot \tan(23^\circ) \]

5. Using a calculator to find \( \tan(23^\circ) \):

   • \( \tan(23^\circ) \approx 0.4245 \)
   • Therefore: \[ LF \approx 47 \cdot 0.4245 \approx 19.94 \text{ feet} \]

6. Rounding to the nearest foot:

   • Thus, \( LF \approx 20 \) feet.

The answer is 20 ft.