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Posted 27 June 2001 - 07:26 PM
Posted 27 June 2001 - 07:44 PM
Posted 28 June 2001 - 04:13 AM
Posted 28 June 2001 - 05:15 AM
Posted 28 June 2001 - 06:56 AM
Posted 29 June 2001 - 05:23 AM
Posted 29 June 2001 - 08:09 AM
Posted 29 June 2001 - 08:20 AM
Posted 29 June 2001 - 08:30 AM
Posted 29 June 2001 - 09:46 AM
quote:
Original post by Wizardry
Keep in mind, however, that using the tangent only works on right triangles. It will not work for any other type of triangle. The problem with that is that you need to know the measure of one of the angles(the 90 degree one).
The law of cosines solution posted earlier is the correct one.
To review:
cos(A) = (b^{2} + c^{2} - a^{2})/(2bc)
Solving for A will give you the angle, no matter what type of triangle you''re dealing with.
Peace out.
Posted 29 June 2001 - 10:24 AM
Posted 29 June 2001 - 04:23 PM
quote:
Original post by BeerNutts
Simple answer (Where Cos-1 = inverse cosine)
Cos(X) = Y
Cos-1(Cos(X)) = Cos-1(Y)
X = Cos-1(Y)
Basically, the inverse cosine of a cosine of an angle is the angle. The opposite is true also.
If you want a deeper answer, then I can answer with many more words.
Nutts
Edited by - BeerNutts on June 29, 2001 5:26:54 PM
Posted 30 June 2001 - 05:25 AM
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