the radius of the earth is approximately 6371 km. if the international space station (ISS) is orbiting 353 km above the earth, find the distance from the ISS to the horizon (x)
the picture is attached any help will be appreciated thaaanks

Question

anonymous · Answer

attachment

anonymous · Answer

|dw:1364777063693:dw|

anonymous · Answer

some geometry. you will get a similarity of triangles.

anonymous · Answer

how would i do this?

anonymous · Answer

$$x^2=(R+h)^2-R^2\
x^2=2Rh+h^2$$

anonymous · Answer

infact, no need for anything

anonymous · Answer

u used the pythagorean theorem?

anonymous · Answer

yep. for triangle OIB

anonymous · Answer

but um im not getting it .

anonymous · Answer

getting what? the answer?

anonymous · Answer

no no like im still confused 
im sorry im so out of it

anonymous · Answer

the line of horizon is tangent to the surface of the earth (a sphere)
any radius drawn to the point of tangent is perpendicular to the tangent. |dw:1364786201261:dw|

anonymous · Answer

you see it now?

anonymous · Answer

$$x^2 +r^2=(r+h)^2$$

anonymous · Answer

we have no values though

anonymous · Answer

yea we do
R = radius of earth
h - height of satel;lite from surface of earth

anonymous · Answer

the radius is half of 6371

anonymous · Answer

ohhh i got it  the height is 353

anonymous · Answer

? but it  states the radius explicitly.

anonymous · Answer

ooh sorry its 6371 thats the radius my bad

anonymous · Answer

yep/

anonymous · Answer

one mor ething the answers r coming out huuuuge is that how its supposed to be?

anonymous · Answer

welll when its all solved i got 2,150

anonymous · Answer

did you take the finl square-root?

anonymous · Answer

that is what it seems to be

anonymous · Answer

provided all units are in "km"
then the final answer is in "km"

anonymous · Answer

this is what i did 
$$x ^{2}+6371^2=(6371+353)^2$$$$x^2+40,589,641=45,212,76$$$$\sqrt{x}^{2} = \sqrt{4,622,535}$$x=2,150