i.
sin(A) = √(1cos^{2}A) = ± 4/5
since A is acute then sin(A) = +4/5
sin(A+B) = √(1cos^{2}(A+B)) = ± 12/13
since both A and B are acute , then A+B is either in first or second quadrants , therefore Sin(A+B) = + 12/13
We have two equations for cos(A+B) and Sin(A+B)
Cos(A+B) = Cos A Cos B  Sin A Sin B => 5/13 = 3/5 Cos B  4/5 Sin B ... eqn I
Sin(A+B) = Sin A Cos B + Cos A Sin B => 12/13 = 4/5 Cos B + 3/5 Sin B ... eqn II
multiply eqn I by 3/4 and add it to eqn II
5*3/(4*13)+12/13 = (9/20 + 4/5) Cos B => Cos B = 33/65 => Sin B = √(1cos^{2}B) = 56/65

ii .
Cos(C) = √(1sin^{2}C) = 5/13 (positive as C is acute)
Sin (A+B+C) = Sin (A+B) Cos C + Cos (A+B) Sin C = 12/13*5/13 + (5/13)*12/13 = 0
iii.
Since Sin(A+B+C) = 0 => A+B+C = 180 or 360
but since A, B and C are acute => their sum is less than 90+90+90 ( i.e. 270)
Therefore A+B+C = 180
Therefore they can form angles of a triangle
2/11/2014

Ahmed F.