Geometry On Earth

a tangent and radius meet at 90 degrees

use Pythagorean theorem 

c= 3959mi + 4.5mi

c= 3963.5mi

a^2 = c^2 - b^2

a^2 = 3963.5^2 - 3959^2

a^2 = 35651.25

a= √35651.25

a= 188.8mi (nearest tenth)

nevermind

can someone answer another question?

no problem

yes me please

A person standing at the top of Mountain Rainier would be approximately 2.7 mi high. The radius of earth is 3959 mi.

What is the distance to the horizon from this point?

Enter your answer as a decimal in the box. Round only your final answer to the nearest tenth.

nevermind

oh ok lol

do we need to find the point at the centre of the earth or the point on the circumfernce of earth?

im not sure i dont think so 

the answer is 146.2

oh ok how did you get the answe?

 The line of sight to the horizon , the radius at the horizon and the radius at the mountain from a right angle triangle, so Pythagoras's theorem gives 

d = sqrt( ((R+h)^2 - R^2) 

= sqrt( 2Rh + h^2) 

= 146.2mi

A person standing at the top of Mountain Rainier would be approximately 2.7 mi high. The radius of earth is 3959 mi.

What is the distance to the horizon from this point?

We can use the Pythagorean Theorem, here 

We have a right triangle with a hypotenuse of 3959 + 2.7  = 3961.7 mi

And the radius of the Earth represents one of the legs...so....the distance to the horizon is the other leg and given by :

√ [ 3961.7^2  - 3959^2 ]   ≈  146.2 mi

thank you both so much!

no problem any time 

There is a very simple formula for this kind of problem:

D = ~Sqrt[1.5 x H], where D =distance in miles, H=Height in feet.

D =~sqrt[1.5 x 2.7 miles x 5,280 feet]

D =~Sqrt[ 21,384]

D=~ 146.23 miles - distance to the horizon.