SOLUTION: A climber is standing at the top of Mount Kazbek, approximately 3.1 mi above sea level. The radius of the Earth is 3959 mi. What is the climber's distance to the horizon?

Question 1083403: A climber is standing at the top of Mount Kazbek, approximately 3.1 mi above sea level. The radius of the Earth is 3959 mi.

What is the climber's distance to the horizon?
Round only your final answer to the nearest tenth. Leave your answer as a decimal.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Start off by drawing the problem out. Make sure you label everything as much as possible. This is one way to do it.

The points A through D are:
A = peak of the mountain where the climber is located
B = base of the mountain (directly below the climber)
C = center of the Earth
D = furthest point (on the horizon) that the climber can see

The segment lengths are:
AD = x
AB = 3.1
BC = 3959
CD = 3959

Using segments AB and BC, we can say
AC = AB + BC
AC = 3.1+3959
AC = 3962.1
assuming ABC is a straight angle.
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Since D is a tangent point, angle ADC is a right angle. Therefore, triangle ADC is a right triangle.

Use the pythagorean theorem to solve for x

(AD)^2 + (CD)^2 = (AC)^2
(x)^2 + (3959)^2 = (3962.1)^2
x^2 + 15673681 = 15698236.41
x^2 + 15673681 - 15673681 = 15698236.41 - 15673681
x^2 = 24555.41
sqrt(x^2) = sqrt(24555.41)
x = 156.701659

The x value is approximate to 6 decimal places.

If you want to round to the nearest tenth, then the accuracy reduces to 1 decimal place.

So we go from 156.701659 to 156.7

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Final Answer: 156.7 miles
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