(Newbie) Triangle ABC

Supermario · **Joined:** Wed, 21 May 2003 07:41:21 UTC **Posts:** 2

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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Mother Goose · **Posted:** Wed, 21 May 2003 08:15:27 UTC

Hey Supermario,

You can use the cosine law!

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

Supermario · **Joined:** Wed, 21 May 2003 07:41:21 UTC **Posts:** 2

A fast and beautiful answer.

Thank's Mother Goose.

Regards,
SuperMario

Jane12 · **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.

bugzpodder · **Joined:** Sun, 4 May 2003 16:04:19 UTC **Posts:** 2906

same diagram,

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

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(Newbie) Triangle ABC

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