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**bigpaully****Member**- Registered: 2005-07-20
- Posts: 4

I am trying to figure out the angle of a triangle with a curved base of 3000km it is part of a circle that has a circumference of 24902km and a radius of 6378km. There is a line that bisects the triangle. Its length is 830km. I need the angle of the triangle that is opposite the 3000km curved side. I wish I could add a picture it would make things easier. The triangle is outside the circle. Think of this problem as a satellite orbiting the earth.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

[I removed the previous reply and banned the user]

It is just a *little* bit hard without a diagram!

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

I caught the fact that you solved it yourself. Well done. But you did have the advantage of a diagram

The measurements sure do sound like this planet.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**bigpaully****Member**- Registered: 2005-07-20
- Posts: 4

here is a link to a diagram of the problem if anyone is still interested

Note: All I was given at the start of the problem was the swath coverage = 3000km (the curved area of the earth the satellite sees) and the altitude of the satellite = 830km.

http://img.photobucket.com/albums/v145/sealcock6/sat_swath_angle.jpg

if you have any questions please reply

*Last edited by bigpaully (2005-07-21 00:30:56)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

OK, well angle ARC is 360° × (3000/24902)

The angle ARE is half that.

The length RE is therefore cos(360° × (3000/24902) /2 ) × 6,378km

RE = cos(21.685°) × 6,378km = 5,927 km

AE = sin(21.685°) × 6,378km = 2,357 km

So distance ED = (6,378-5,927) + 830 = 1,281 km

We now know AE and ED, so the angle from that is:

tan-¹(2,357/1,281) = 61.5°, and you need to double that to 123° --> ANSWER = 123°

There may be an easier way to get there, I just went about soloving triangles till I got there.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**bigpaully****Member**- Registered: 2005-07-20
- Posts: 4

right on. thats what i got well 122.88 degree. i needed to be a little more accurate.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

Well, if we both got a similar answer it MUST be right.

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**Zach****Member**- Registered: 2005-03-23
- Posts: 2,075

Either that or you're both wrong.

Boy let me tell you what:

I bet you didn't know it, but I'm a fiddle player too.

And if you'd care to take a dare, I'll make a bet with you.

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