Question
Your town’s public library is building a new wheelchair ramp to its entrance. By law, the maximum angle of incline for the ramp is 4.76°. The ramp will have a vertical rise of 2 ft. What is the shortest horizontal distance that the ramp can span?
(1 point)
Responses
15.3 ft.
15.3 ft.
3.5 ft.
3.5 ft.
24.0 ft.
24.0 ft.
9.5 ft.
Find the exact value of csc θ if cotangent theta equals negative 2 and the terminal side of θ lies in quadrant II.
(1 point)
Responses
cosecant theta equals start root 5 end root
Image with alt text: cosecant theta equals start root 5 end root
cosecant theta equals negative start root 5 end root
Image with alt text: cosecant theta equals negative start root 5 end root
cosecant theta equals negative start fraction start root 5 end root over 2 end fraction
Image with alt text: cosecant theta equals negative start fraction start root 5 end root over 2 end fraction
cosecant theta equals start fraction start root 5 end root over 2 end fraction
(1 point)
Responses
15.3 ft.
15.3 ft.
3.5 ft.
3.5 ft.
24.0 ft.
24.0 ft.
9.5 ft.
Find the exact value of csc θ if cotangent theta equals negative 2 and the terminal side of θ lies in quadrant II.
(1 point)
Responses
cosecant theta equals start root 5 end root
Image with alt text: cosecant theta equals start root 5 end root
cosecant theta equals negative start root 5 end root
Image with alt text: cosecant theta equals negative start root 5 end root
cosecant theta equals negative start fraction start root 5 end root over 2 end fraction
Image with alt text: cosecant theta equals negative start fraction start root 5 end root over 2 end fraction
cosecant theta equals start fraction start root 5 end root over 2 end fraction
Answers
GPT-4o mini
To solve the first problem about the wheelchair ramp:
Given:
- Maximum angle of incline, \( \theta = 4.76° \)
- Vertical rise, \( h = 2 \) ft
We can use the tangent function, which relates the angle of incline to the vertical rise and horizontal distance. The tangent of the angle is given by:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{d}
\]
where \( d \) is the horizontal distance. Rearranging the formula gives us:
\[
d = \frac{h}{\tan(\theta)}
\]
Calculating \( d \):
1. Find \( \tan(4.76°) \). You can use a calculator for the tangent of 4.76 degrees:
\[
\tan(4.76°) \approx 0.0831
\]
2. Plug in the values:
\[
d = \frac{2}{\tan(4.76°)} \approx \frac{2}{0.0831} \approx 24.1 \text{ ft}
\]
Since the options provided are approximate, the shortest horizontal distance that the ramp can span is:
**24.0 ft.**
---
For the second problem regarding \( \csc(\theta) \):
Given:
- \( \cot(\theta) = -2 \)
- Quadrant II
From the definition of cotangent:
\[
\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = -2
\]
This implies that:
\[
\frac{\cos(\theta)}{\sin(\theta)} = -2 \Rightarrow \cos(\theta) = -2\sin(\theta)
\]
Let \( \sin(\theta) = y \) and \( \cos(\theta) = x \), we can express:
\[
x = -2y
\]
Using the Pythagorean identity \( x^2 + y^2 = 1 \):
\[
(-2y)^2 + y^2 = 1
\]
\[
4y^2 + y^2 = 1
\]
\[
5y^2 = 1 \Rightarrow y^2 = \frac{1}{5} \Rightarrow y = \frac{1}{\sqrt{5}} \text{ (since y is positive in Quadrant II)}
\]
Now, we find \( x \):
\[
x = -2y = -2\left(\frac{1}{\sqrt{5}}\right) = -\frac{2}{\sqrt{5}}
\]
To find \( \csc(\theta) \):
\[
\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{y} = \frac{1}{\frac{1}{\sqrt{5}}} = \sqrt{5}
\]
Thus, the exact value of \( \csc(\theta) \) is:
**cosecant theta equals start root 5 end root**
Given:
- Maximum angle of incline, \( \theta = 4.76° \)
- Vertical rise, \( h = 2 \) ft
We can use the tangent function, which relates the angle of incline to the vertical rise and horizontal distance. The tangent of the angle is given by:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{d}
\]
where \( d \) is the horizontal distance. Rearranging the formula gives us:
\[
d = \frac{h}{\tan(\theta)}
\]
Calculating \( d \):
1. Find \( \tan(4.76°) \). You can use a calculator for the tangent of 4.76 degrees:
\[
\tan(4.76°) \approx 0.0831
\]
2. Plug in the values:
\[
d = \frac{2}{\tan(4.76°)} \approx \frac{2}{0.0831} \approx 24.1 \text{ ft}
\]
Since the options provided are approximate, the shortest horizontal distance that the ramp can span is:
**24.0 ft.**
---
For the second problem regarding \( \csc(\theta) \):
Given:
- \( \cot(\theta) = -2 \)
- Quadrant II
From the definition of cotangent:
\[
\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = -2
\]
This implies that:
\[
\frac{\cos(\theta)}{\sin(\theta)} = -2 \Rightarrow \cos(\theta) = -2\sin(\theta)
\]
Let \( \sin(\theta) = y \) and \( \cos(\theta) = x \), we can express:
\[
x = -2y
\]
Using the Pythagorean identity \( x^2 + y^2 = 1 \):
\[
(-2y)^2 + y^2 = 1
\]
\[
4y^2 + y^2 = 1
\]
\[
5y^2 = 1 \Rightarrow y^2 = \frac{1}{5} \Rightarrow y = \frac{1}{\sqrt{5}} \text{ (since y is positive in Quadrant II)}
\]
Now, we find \( x \):
\[
x = -2y = -2\left(\frac{1}{\sqrt{5}}\right) = -\frac{2}{\sqrt{5}}
\]
To find \( \csc(\theta) \):
\[
\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{y} = \frac{1}{\frac{1}{\sqrt{5}}} = \sqrt{5}
\]
Thus, the exact value of \( \csc(\theta) \) is:
**cosecant theta equals start root 5 end root**