There we go. I knew this was easy. What you've done is attack this problem the same way my text does, but with a much more straightforward approach that I like. For example, my text would create an equation that would lead to finding the direction of the boat relative to the water thus: Vc = 4i (water current) Vw = ai + bj, so ||Vw|| = sqrt(a^2 + b^2) Vg = Vw + Vc (from the diagram above) kj = ai + bj + 4i, where k is the magnitude of the vector Vg, so kj = (a +4)i + bj, noting that for this equation to be true, the coefficients of the i and j components must be equal on both sides. Solving, a = -4 and b = sqrt(209), so Vw = -4isqrt(209)j and Vg = kj = sqrt(209)j We then find angle theta between Vw and Vg: COS theta = the dot product of Vw and Vg divided by ||VW|| ||VG|| = .9638, or 15.5 degrees. It's functional, but I wouldn't consider it efficient when I can just draw a right triangle and be done with it. At least now I can, so thanks! I also see why my first attempt flubbed..