Math Problems

I found the triangle DCP:
The bottom, DC, is 3QC. (DQ is 2QC, and QC is QC)
The right, PC, is 3PB. (Given)

The right side is 4 PB (3PB + PB)

Both sides are equal to k, so 3QC = 4PB

QC = 4/3 * PB

tangent (3PB/3QC) = tan(PB/QC) = tan(4/3*QC/QC) = tan(3/4) ~= 36.87 degrees.

You know an angel and a side, I think that's enough to break down the rest of the triangle's sides / area. I haven't done this in two years, so iunno.

Edit: Nope, you need another side. Rawr. I'll get back to this.
Leaving this here for reference. http://gcse.wikia.com/wiki/Area_of_a_triangle