If the sides of a triangle are in AP, and the greatest angle of the triangle is double the smallest angle, the ratio of the sides of the triangle is

Solution:

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 angle and C is the largest angle. It is given that C = 2A. 

Thus, the angles of the triangle are A, 2A and  $\mathrm{\pi }$-3A. 

Applying the law of sines, we obtain 

$\begin{array}{rl} & \frac{\mathrm{a}-\mathrm{d}}{\sin \mathrm{A}}=\frac{\mathrm{a}}{\sin (\mathrm{\pi }-3\mathrm{A})}=\frac{\mathrm{a}+\mathrm{d}}{\sin 2\mathrm{A}}\\ \Rightarrow & \frac{\mathrm{a}-\mathrm{d}}{\sin \mathrm{A}}=\frac{\mathrm{a}}{\sin 3\mathrm{A}}=\frac{\mathrm{a}+\mathrm{d}}{\sin 2\mathrm{A}}\\ \Rightarrow & \frac{\mathrm{a}-\mathrm{d}}{\sin \mathrm{A}}=\frac{\mathrm{a}}{3\sin \mathrm{A}-4{\sin }^3\mathrm{A}}=\frac{\mathrm{a}+\mathrm{d}}{2\sin \mathrm{Acos}\mathrm{A}}\\ \Rightarrow & \frac{\mathrm{a}-\mathrm{d}}{1}=\frac{\mathrm{a}}{3-4{\sin }^2\mathrm{A}}=\frac{\mathrm{a}+\mathrm{d}}{2\cos \mathrm{A}}\\ \Rightarrow & 3-4{\sin }^2\mathrm{A}=\frac{\mathrm{a}}{\mathrm{a}-\mathrm{d}}\text{and}2\cos \mathrm{A}=\frac{\mathrm{a}+\mathrm{d}}{\mathrm{a}-\mathrm{d}}\\ \Rightarrow & 4{\cos }^2\mathrm{A}-1=\frac{\mathrm{a}}{\mathrm{a}-\mathrm{d}}\text{and}2\cos \mathrm{A}=\frac{\mathrm{a}+\mathrm{d}}{\mathrm{a}-\mathrm{d}}\end{array}$

$\Rightarrow {\left(\frac{\mathrm{a}+\mathrm{d}}{\mathrm{a}-\mathrm{d}}\right)}^2-1=\frac{\mathrm{a}}{\mathrm{a}-\mathrm{d}}\Rightarrow \mathrm{a}=5\mathrm{d}$

Thus, the sides of the triangle are a-d, a, a + d    i.e. 4d, 5d, 6d.

Hence, the ratio of the sides of the triangle is 4d :5d: 6d     i.e. 4:5:6.