1. ## 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
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.

Any ideas?
Thanks

2. ## 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!

3. ## 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

4. ## Re: Law of Tangents

I'm gonna call it a night

5. ## 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.

6. ## Re: Law of Tangents

Haha no way, I couldn't.. I give up :/

7. ## 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