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A climber stands on a mountain that is 1.55 mi above the surface of the earth. The earth has a radius of 3959 mi. What is the climber's distance from the horizon? Enter your answer, rounded to the nearest mile, in the box.

User Ram Pukar
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A right triangle is made, where the distance between the climber and the horizon is one of the legs, the Earth's radius is another leg and the distance from the climber to the center of the Earth is the hypotenuse. Applying the Pythagorean theorem we get:


\begin{gathered} (3959+1.55)^2=3959^2+x^2 \\ 3960.55^2=3959^2+x^2 \\ 15685956.3=15673681+x^2 \\ 15685956.3-15673681=x^2 \\ 12275.3=x^2 \\ \sqrt[]{12275.3}=x \\ 111\text{ mi = x} \end{gathered}

A climber stands on a mountain that is 1.55 mi above the surface of the earth. The-example-1
User Dlaor
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