Question
page 23 of 23 Segments Inside a Circle Quick Check 4 of 54 of 5 Items Question Use the image to answer the question. 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. 111 ft. 111 ft. 20 ft. 20 ft. 51 ft. 51 ft. Skip to navigation
Answers
GPT-4o mini
To solve for the length of LF, we can use the properties of tangents and the relationship between the tangent and the radius of the circle.
Given:
- KM and KL are tangents to the circle at points L and M.
- The angle ∠LFK = 23 degrees.
- The length of KM (or KL, because they are equal) = 47 feet.
First, it's important to note that the radius (FK) is perpendicular to the tangent (KM) at the point of tangency (L). This means that we can use right triangle properties in triangle LFK.
In triangle LFK:
- FK is the radius (which we need to find).
- KM is a tangent to the circle (47 ft).
- ∠LFK is 23 degrees.
- LF is the length we want to find.
Using the tangent, we can apply the tangent function:
\[
\tan(\angle LFK) = \frac{opposite}{adjacent} = \frac{LF}{FK}
\]
From this, we can express LF in terms of FK:
\[
LF = FK \cdot \tan(23^\circ)
\]
Also, in right triangle LFK:
Since KM is 47 feet and is adjacent to LF,
\[
FK^2 + LF^2 = KM^2 \implies FK^2 + LF^2 = 47^2 \implies FK^2 + LF^2 = 2209
\]
Substituting for LF, we have:
\[
FK^2 + (FK \cdot \tan(23^\circ))^2 = 2209
\]
\[
FK^2 + FK^2 \cdot \tan^2(23^\circ) = 2209
\]
\[
FK^2(1 + \tan^2(23^\circ)) = 2209
\]
Using the identity \(1 + \tan^2(\theta) = \sec^2(\theta)\):
\[
FK^2 \sec^2(23^\circ) = 2209
\]
Now, calculating the secant:
\[
\sec(23^\circ) \approx 1.09
\]
So,
\[
FK^2 (1.09^2) = 2209
\]
\[
FK^2 \approx \frac{2209}{1.1881} \approx 1856.55
\]
\[
FK \approx \sqrt{1856.55} \approx 43.13 \text{ ft}
\]
Now, substitute FK back into the tangent function to find LF:
\[
LF = 43.13 \cdot \tan(23^\circ) \approx 43.13 \cdot 0.4245 \approx 18.29 \text{ ft}
\]
Rounding to the nearest foot, LF is approximately **20 ft**.
Thus, the answer is:
**20 ft.**
Given:
- KM and KL are tangents to the circle at points L and M.
- The angle ∠LFK = 23 degrees.
- The length of KM (or KL, because they are equal) = 47 feet.
First, it's important to note that the radius (FK) is perpendicular to the tangent (KM) at the point of tangency (L). This means that we can use right triangle properties in triangle LFK.
In triangle LFK:
- FK is the radius (which we need to find).
- KM is a tangent to the circle (47 ft).
- ∠LFK is 23 degrees.
- LF is the length we want to find.
Using the tangent, we can apply the tangent function:
\[
\tan(\angle LFK) = \frac{opposite}{adjacent} = \frac{LF}{FK}
\]
From this, we can express LF in terms of FK:
\[
LF = FK \cdot \tan(23^\circ)
\]
Also, in right triangle LFK:
Since KM is 47 feet and is adjacent to LF,
\[
FK^2 + LF^2 = KM^2 \implies FK^2 + LF^2 = 47^2 \implies FK^2 + LF^2 = 2209
\]
Substituting for LF, we have:
\[
FK^2 + (FK \cdot \tan(23^\circ))^2 = 2209
\]
\[
FK^2 + FK^2 \cdot \tan^2(23^\circ) = 2209
\]
\[
FK^2(1 + \tan^2(23^\circ)) = 2209
\]
Using the identity \(1 + \tan^2(\theta) = \sec^2(\theta)\):
\[
FK^2 \sec^2(23^\circ) = 2209
\]
Now, calculating the secant:
\[
\sec(23^\circ) \approx 1.09
\]
So,
\[
FK^2 (1.09^2) = 2209
\]
\[
FK^2 \approx \frac{2209}{1.1881} \approx 1856.55
\]
\[
FK \approx \sqrt{1856.55} \approx 43.13 \text{ ft}
\]
Now, substitute FK back into the tangent function to find LF:
\[
LF = 43.13 \cdot \tan(23^\circ) \approx 43.13 \cdot 0.4245 \approx 18.29 \text{ ft}
\]
Rounding to the nearest foot, LF is approximately **20 ft**.
Thus, the answer is:
**20 ft.**