# Triangle with median and altitude

1. Nov 29, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
In triangle ABC with A as obtuse angle, AD and AE are median and altitude respectively. If BAD = DAE=EAC, then sin^3(A/3)cos(A/3) equals

2. Relevant equations
3. The attempt at a solution

CE = a/2. Let DE = x. Then BD = a/2 - x.
Let AE = p, AD = q.

$\cos \frac{A}{3} = \dfrac{q^2+c^2-(a/2 -x)^2}{2cq}$

I can also write cos A/3 for other two triangles using the same approach but I don't know which one to use in the final expression. Also it will contain p and q which is unknown. I have no idea what sin(A/3) would be. This seems really complicated.

2. Nov 29, 2013

### tiny-tim

hi utkarshakash!
why use such an awkward triangle?

there are three right-angled triangles …

use them!

3. Nov 29, 2013

### utkarshakash

How could I miss them?:tongue2: Thanks a lot.