Triangle with median and altitude

[utkarshakash]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

For ΔADB

\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.

 

[tiny-tim]

hi utkarshakash!

For ΔADB

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

why use such an awkward triangle?

there are three right-angled triangles …

use them! ​

 

[utkarshakash]

hi utkarshakash!


why use such an awkward triangle?

there are three right-angled triangles …

use them! ​

How could I miss them? Thanks a lot.