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 Post subject: (Newbie) Triangle ABC
PostPosted: Wed, 21 May 2003 07:56:38 UTC 
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Joined: Wed, 21 May 2003 07:41:21 UTC
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Hi dear math-friends,

My question is too easy, sorry... :| I'm lawyer and forgot trignometry a long time ago. :cry: I need it again to some programming bitmap rotation.


Imagine a triangle ABC.
I know the 3 sides (AB, AC, BC).

I would like to get the A angle. What is the formula?


:)
Thank you very much!
[[]]


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 Post subject:
PostPosted: Wed, 21 May 2003 08:15:27 UTC 
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Joined: Fri, 2 May 2003 14:09:58 UTC
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Location: In the pub, teaching Corneo how to pick up chicks...
Hey Supermario,

You can use the cosine law!

Image

in your case, if you have AB, AC, and BC, to calculate the angle A, you would:

Let A be the angle
BC^2 = AC^2 + AB^2 - 2(AC)(AB)cosA\\\\
\frac{BC^2 - AC^2 - AB^2}{ -2(AC)(AB)} = cosA \\\\
A = cos^{-1}(\frac{BC^2 - AC^2 - AB^2}{ -2(AC)(AB)})\\

I'm assuming that this is for any triangle. If you have a right triangle, you can always use the trig ratios

Hope it turns out okay ;)


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 Post subject: Thank's
PostPosted: Wed, 21 May 2003 08:53:51 UTC 
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Joined: Wed, 21 May 2003 07:41:21 UTC
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:)

A fast and beautiful answer.

Thank's Mother Goose.

Regards,
SuperMario


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 Post subject: Could you do one for law of sines please?
PostPosted: Thu, 22 May 2003 21:25:59 UTC 
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Joined: Tue, 20 May 2003 01:48:07 UTC
Posts: 28
That was the best illustration I have ever seen of the law of cosines. I wonder if you could do one for the law of sines. I really need this stuff done in a clear way. :o


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 Post subject:
PostPosted: Thu, 22 May 2003 21:33:29 UTC 
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Joined: Sun, 4 May 2003 16:04:19 UTC
Posts: 2906
same diagram,

\[
\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}
\]

_________________
Has anyone noticed that the below is WRONG? Otherwise this statement would be true:
-1\cong1\pmod{13}
i\cong5 \pmod{13} where
i^2=-1


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