# Gibb's Hill Lighthouse

Discussion in 'Algebra' started by nycmathguy, Jul 10, 2022.

1. ### nycmathguy

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College Algebra
Chapter 1/Section 3

Can you work this one out for me?

nycmathguy, Jul 10, 2022

2. ### MathLover1

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here is sketch for you to try do it yourself

to verify the accuracy of these statements use the facts you know about the Pythagorean Theorem, that the radius of Earth is 3960 miles, and 1 mile = 5280 ft

MathLover1, Jul 10, 2022
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3. ### nycmathguy

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Thanks. I will tackle this myself some other day. Brain dead.

nycmathguy, Jul 11, 2022
4. ### nycmathguy

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I can't figure it out even with the information and pictures given. Can you set up the equations needed for me to answer the question? What is the question asking for Exactly? I noticed that you left several numbers given in the application out of your diagram. Just set up the equation(s) needed for me to correctly answer the question.

Thanks

nycmathguy, Jul 11, 2022
5. ### MathLover1

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Using the Pythagorean theorem the first statement was verified using 3960

d^2+3960^2=(3960+(362/5280))^2
d^2=(3960+(362/5280))^2-(3960)^2
d=23.3
So the lighthouse cannot be seen from a distance of 26 miles.

The brochure further states that ships 40 miles away can see the light and planes flying at 10,000 feet can See it 120 miles away. Verify the accuracy of these statements. What Assumption did the brochure make about the height of the ship? (Use 3960 as the radius of the earth).

hypothenuse=10,000 feet +3960 miles=3962 miles
first find s
3960+s/5280=sqrt(15683200)
s/5280=3960.202 -3960
s/5280=0.202
s=0.202miles

then
120^2 +(3960+s/5280)^2=3962^2
120^2 +(3960+0.202)^2=3962^2
15697600=15697444 -> approximately same, so result is confirmed

MathLover1, Jul 11, 2022
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