Height of a Tetrahedron (Pyramid) - HELP

Question

ABCD is a tetrahedron whose horizontal base ABC is an equilateral triangle. The angle between each pair of slant edges is θ where tan θ = 5/12, and the length of these edges is a. Find the height of D above ABC.

The Answer I keep getting is h = (a)sqrt(202/208), whereas the answer in the book is h = (a)sqrt(37/39). I have shown my working below:

 

http://i.stack.imgur.com/6fRqE.jpg

 

Thanks

 

EDIT: SOLVED, Incorrectly Assumed that the centre of the equilateral triangle fell at the midpoint of the midline. It in fact falls 2/3rds along it. Solution here: http://oi60.tinypic.com/23upug.jpg.

Answer 1

Starting with tan(a/2) = sqrt((1-cos(a))/(1+cos(a))) might be a little quicker