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Jessica sets up a volleyball net in her backyard. To secure the net in place, she extends a rope from the top of each of the tw...Asked by PJ
Jessica sets up a volleyball net in her backyard. To secure the net in place, she extends a rope from the top of each of the two poles at a diagonal to the ground. Once the poles are placed in the ground, they are 7 feet tall; the rope she uses is 9 feet long. Using the inverse of sine, what is the approximate angle formed between the ground and the rope?(1 point)
Responses
90°
90°
45°
45°
51°
51°
39°
Responses
90°
90°
45°
45°
51°
51°
39°
Answers
Answered by
GPT-4o mini
To find the angle formed between the ground and the rope using the inverse sine function, we can use the sine definition in a right triangle context. The setup is as follows:
- The height of the pole (opposite side) = 7 feet.
- The length of the rope (hypotenuse) = 9 feet.
We can use the formula for sine:
\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
Substituting the given values:
\[
\sin(\theta) = \frac{7}{9}
\]
To find the angle θ, we take the inverse sine (arcsin) of \( \frac{7}{9} \):
\[
\theta = \arcsin\left(\frac{7}{9}\right)
\]
Calculating \( \arcsin\left(\frac{7}{9}\right) \):
Using a calculator:
\[
\theta \approx 51.06°
\]
Therefore, rounding to the nearest whole number, the approximate angle formed between the ground and the rope is:
**51°**
So the correct response is:
**51°**
- The height of the pole (opposite side) = 7 feet.
- The length of the rope (hypotenuse) = 9 feet.
We can use the formula for sine:
\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
Substituting the given values:
\[
\sin(\theta) = \frac{7}{9}
\]
To find the angle θ, we take the inverse sine (arcsin) of \( \frac{7}{9} \):
\[
\theta = \arcsin\left(\frac{7}{9}\right)
\]
Calculating \( \arcsin\left(\frac{7}{9}\right) \):
Using a calculator:
\[
\theta \approx 51.06°
\]
Therefore, rounding to the nearest whole number, the approximate angle formed between the ground and the rope is:
**51°**
So the correct response is:
**51°**
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