Law of Tangents

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January 16th 2013, 12:48 PM
Imonars

Law of Tangents

Hey MHF, today in class we took Melloweid's formula, then the teacher gave us this equation and only two students were able to solve it (Headbang)
I'm 100% sure I'm going to have to use Melloweid's formula to derive it.

Law of Tangents For any triangle, derive the Law of Tangents.
http://i46.tinypic.com/smrqt0.png

Any ideas?
Thanks
January 16th 2013, 01:18 PM
Barioth

Re: Law of Tangents

Hi the Melloweird's formula are given there

Mollweide's formula - Wikipedia, the free encyclopedia

if you take the second one and divide it by the firt one you end up with

$\frac{(a-b)}{(a+b} = \tan{(\frac{1}{2}(A-B)}\tan{(\frac{\theta}{2})}$
we know that
$A+B = 180 - \theta$
so
$\tan{(\frac{\theta}{2})} = \tan{(90 -\frac{A+B}{2})}$

The tricky part is the find out Why in the world
$\tan{(90 -\frac{A+B}{2})} = \arctan{(\frac{A+B}{2}})$

But this is the funny part so I leave it to you!
January 16th 2013, 01:43 PM
Imonars

Re: Law of Tangents

Lol, I will give it a thought and see how it goes, if I find it I will post the results. If not, then know that I have failed :)
January 16th 2013, 03:41 PM
Imonars

Re: Law of Tangents

I'm gonna call it a night (Sleepy)
January 16th 2013, 04:06 PM
Barioth

Re: Law of Tangents

dang I think I made a mistake there! Hold on i'll change few thing..

there it should be fine. Sorry about that.
January 17th 2013, 02:09 AM
Imonars

Re: Law of Tangents

Haha no way, I couldn't.. I give up :/
January 17th 2013, 07:40 AM
Barioth

Re: Law of Tangents

Here is how you solve it

sin(a-b) = sin a*cos b - sin b*cos a

Let a = pi/2 and b = x = (A+B)/2

sin ( /2 - x) = sin /2*cos x - sin x* cos /2

since sin pi/2 = 1 and cos pi/2 = 0

sin ( pi/2 - x) = 1*cos x - sin x*0

sin ( pi/2 - x) = cos x

note: I personaly just write down my trigo circle and see it eazier from there.

Have a good day

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