# Law of Tangents

• Jan 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
• Jan 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

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

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

But this is the funny part so I leave it to you!
• Jan 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 :)
• Jan 16th 2013, 03:41 PM
Imonars
Re: Law of Tangents
I'm gonna call it a night (Sleepy)
• Jan 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.
• Jan 17th 2013, 02:09 AM
Imonars
Re: Law of Tangents
Haha no way, I couldn't.. I give up :/
• Jan 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