Two astronomers A, B located in different regions A: 17° E, 46 ° 27 'N, B: 17° E, 22 ° 34 'S observe an asteroid at the moment it is exactly above the meridian and is in position C which is a distance a from A and b from B.

Both observe from sea level and astronomer A sees the asteroid 25 ° south of his zenith, while astronomer B sees the asteroid 45 ° north of his zenith.

Consider the radius of Earth equal to 6400km.

Calculate the distance of this asteroid from the center of the Earth.

My work: We can create two triangles for example triangle AEC and BEC(Where E is the center of Earth) and from the data we have one angle for each (EAC = 180-25=155 and EBC=180-45=135) and one known shared side which is the radius of Earth. This is how far I have gone I suspect using law of sines and cosines to find some relations and do a system of equations but whenever I try and solve the system the numbers don't make sense.

Any ideas on how to continue with the soution will be appreciated. Thanks in advance for your help.


1 Answer 1


Draw a diagram.

You get a quadrilateral EACB. You should be able to determine the four angles in the quadrilateral.

Draw the diagonal EC, which is the length to be determined.

You know that ACE + BCE = ACB (the known angle at the asteroid)

And you can use the law of sines twice, once in each of the triangles EAC and EBC

These should give you three equations, with three unknowns: length EC, and angles ACE and BCE. Given three equations and three unknowns, a solution should be possible. You should find the distance is a little more than the Earth-moon distance.

Note that angles at the centre made by the diagonal: AEC and BEC are not 46° 27' and 22° 34', but the sum of those angles is the sum of 46° 27' + 22° 34'


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