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**seren****Member**- Registered: 2005-12-10
- Posts: 2

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I'm fairly sure that it can be worked out. Is the third point at the equator or just at an unknown place?

Why did the vector cross the road?

It wanted to be normal.

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**seren****Member**- Registered: 2005-12-10
- Posts: 2

yep at equator - just my dodgy drawing :$

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

In that case, it is definitely possible.

70S is around 5990 km below the equator and 80N is around 6280km above it.

If the 70S point is 6370km away from the equator and 5990km below it, then it must be √(6370² - 5990²) = 2167km to the left of the middle point.

Similarly, if the 80N point is 6370km away from the equator and 6280km above it, then it must be √(6370² - 6280²) = 1067km away from it.

Therefore, the 80N and 70S points are 5990 + 6280 = 12270km away from each other by latitude and 2167 - 1067 = 1100km away by longitude.

That means that they are √(12270² + 1100²) ≈ **12320km away from each other.**

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

Careful ... it is a curved surface, so Pythagoras may not work ...

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

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**irspow****Member**- Registered: 2005-11-24
- Posts: 457

If you are looking for the distance along the surface of the earth.

Since 80° N above the equator and 70° S below the equator total 150° for the interior angle. We can use:

s = rθ in rads or s = r π θ ÷ 180 in degrees (s = arc length)

s = 6370(150π):div180 = 16676.6 km

If you want straight line distance between the two points then:

Since the interior angle of the apex is 150° and the radius is estimated to be 6370 we can use the law of sines. The radius is constant and 180° - 150° = 30° tells us that the other angles are both 15°

x/sin 150° = 6370/sin 15°

So x = 6370 sin 150° / sin 15° ≈ 12305.9 km

Hope that helps.

*Last edited by irspow (2005-12-10 12:02:38)*

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