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A streetlight post needs some extra support for potential bad weather and high winds. The city wants to install metal support b...Asked by Dino cat
A streetlight post needs some extra support for potential bad weather and high winds. The city wants to install metal support brackets on the light post. Each bracket will be braced into the ground and will then attach to a point on the pole that is 4.5 ft. from the ground. Each bracket is 6.5 ft. long. How far from the base of the pole should each bracket be braced in the ground? Round to the nearest tenth. (1 point)
22.0 ft
4.7 ft
7.9 ft
62.5 t
22.0 ft
4.7 ft
7.9 ft
62.5 t
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Answered by
GPT-4o mini
To find out how far from the base of the pole each bracket should be braced in the ground, we can use the Pythagorean theorem. The triangle formed by the height of the attachment point on the pole (4.5 ft), the distance from the base of the pole (let's call it \( x \), in ft), and the length of the bracket (6.5 ft) represents a right triangle.
According to the Pythagorean theorem:
\[
a^2 + b^2 = c^2
\]
where:
- \( a \) is the height from the ground to the attachment point (4.5 ft),
- \( b \) is the distance from the base of the pole (the value we need to find, \( x \)),
- \( c \) is the length of the bracket (6.5 ft).
Plugging in the values, we have:
\[
(4.5)^2 + x^2 = (6.5)^2
\]
Calculating the squares:
\[
20.25 + x^2 = 42.25
\]
Now, solve for \( x^2 \):
\[
x^2 = 42.25 - 20.25
\]
\[
x^2 = 22
\]
Now take the square root:
\[
x = \sqrt{22} \approx 4.69
\]
Rounding to the nearest tenth, we get:
\[
x \approx 4.7 \text{ ft}
\]
Thus, each bracket should be braced approximately **4.7 ft** from the base of the pole.
According to the Pythagorean theorem:
\[
a^2 + b^2 = c^2
\]
where:
- \( a \) is the height from the ground to the attachment point (4.5 ft),
- \( b \) is the distance from the base of the pole (the value we need to find, \( x \)),
- \( c \) is the length of the bracket (6.5 ft).
Plugging in the values, we have:
\[
(4.5)^2 + x^2 = (6.5)^2
\]
Calculating the squares:
\[
20.25 + x^2 = 42.25
\]
Now, solve for \( x^2 \):
\[
x^2 = 42.25 - 20.25
\]
\[
x^2 = 22
\]
Now take the square root:
\[
x = \sqrt{22} \approx 4.69
\]
Rounding to the nearest tenth, we get:
\[
x \approx 4.7 \text{ ft}
\]
Thus, each bracket should be braced approximately **4.7 ft** from the base of the pole.
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