A, B, p
Start with drawing a triangle AB'C' with two given angles A and B. However you do this, the triangle you
obtain will be similar to the triangle ABC we have to find. On cc, starting with the point A, measure the perimiter
of AB'C' (point P) and 2p (point Q). Connect P and C'. Through Q, draw a line parallel to PC' that intersects
bb at the point C. From C draw a line parallel to C'B' to obtain B.
A remark is in order here. Two angles completely define a triangle's shape. To determine a triangle
uniquely one has to fix its size. Since in similar triangles all linear elements are in the same ratio,
any linear element in a triangle uniquely determines that triangle. Therefore, the above solution will work
as well in other cases. For example, when we are given A,B, and, say, lc or HaMb.