The sides of a triangle are in AP and its area is 3 / 5 of an equilateral triangle of the same perimeter prove that the lebgth of the sides are 3:5:7, and find the gratest angle of the triangle.

Let sides be a-d,a,a+d therefore s=3a/2 Area = √3a/2(3a/2 - a)(3a/2 -a+d)(3a/2 -a-d)= 3/5×√3/4 a^2 On solving we get a/4×√a^2 -4d^2 = 3/5×a^2/4 Further 16a^2=100d^2 =4a=10d or 2a=5d hence ratio -a-d:a:a+d = 3:5:7 (a-2a/5:a:a+2a/5) And greatest angle = (3a)^2+(5a)^2 - (7a)^2/2×3a×5a = cosA On solving cosA= -1/2 hence A = 120°