Upwards tangent vector of triangle in 3D space
I have a triangle in 3D cartesian space, it forms a surface. I have a normal vector of that surface. What I want to find out, is a vector tangent to that surface, which points the most "upwards". (The orange one on image, forgive my paint skills) 
Solution
Let one triangle edge vector is A. Get perpendicular vector in the plane
P = N x A
and normalize P and A
p = P / len(P)
a = A / len(A)
Any unit vector in the plane is combination of these base vectors
v = p * cos(t) + a * sin(t) (1)
We want that Z-component of v to be maximal (as far as I understand most "upwards")
vz = pz * cos(t) + az * sin(t) (2)
has extremum when it's derivative by t is zero
0 = (pz * cos(t) + az * sin(t))' = -pz * sin(t) + az * cos(t)
tan(t) = az / pz
t = atan2(az , pz)
put t values into (1) and get needed vector v
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