1. We can sovle the first one using the Law of Cosines.......we need to find the angle opposite the greatest side
42^2 = 34^2 + 28^2 - 2 (34)(28)cos (theta)
cos (theta) = [ 42^2 - 34^2 - 28^2] / [ -2 * 34 * 28 ]
arccos ( [ 42^2 - 34^2 - 28^2] / [ -2 * 34 * 28 ] ) = theta ≈ 84.7°
Since the greatest angle lies opposite the greatest side, this triangle is acute
2. Since angle B = 45, then so does angle C.......and if two angles of a triangle are equal, then the sides opposite those angles are also equal......so CA = BA
