If the sides of a triangle are in A.P. and the greatest angle of the triangle is double the smallest angle, then the ratio of the sides of the triangle is
4:5:6
Let the sides of the triangle be a-d,a and a+d with a>d>0. Clearly, a-d is the smallest side and a+d is the largest side. So A is the smallest and C is the largest angle. Then c=2A. Thus the angles of the triangle are, Applying the law of sines we have
a−dsinA=asin(π−3A)=a+dsin2A
⇒a−dsinA=asin3A=a+dsin2A⇒a−dsinA=a3sinA−4sin3A=a+d2sinAcosA
⇒a−d1=a3−4sin2A=a+d2cosA⇒3−4sin2A=aa−d
2cosA =a+da−d
⇒4cos2A−1=aa−d
2cosA=a+da−d
⇒(a+da−d)2−1=aa−d
⇒a=5d
Thus, the sides of the triangle are 4d, 5d, 6d. Hence, their ratio is 4:5:6