Question

The sides of a triangle are in $A.P$, and the greatest angle is double of smallest angle. Prove that the ratio of its sides is $4:5:6$.

Answer 1

Let the sides be $a-d,a,a+d$
It is understood that $a>d>0$ and from the figure $\angle C$ is greatest and $\angle A$ is smallest.
By given condition $C=2A$
And hence $B=\pi (A+C)=\pi -3A$
Hence by sine rule we have
$\frac{a+d}{\sin C}=\frac{a-d}{\sin A}=\frac{a}{\sin B}$
or $\frac{a+d}{\sin 2A}=\frac{a-d}{\sin A}=\frac{a}{\sin (\pi -3A)}$
$\therefore 2\cos A=\frac{a+d}{a-d}$ and $\frac{a}{a-d}=\frac{\sin 3A}{\sin A}$
$\therefore \frac{a}{a-d}=3-4{\sin }^{2}A=3-4+{\left(2\cos A\right)}^2$
$\therefore \frac{a}{a-d}=1+{\left(\frac{a+d}{a-d}\right)}^2=\frac{4ad}{{(a-d)}^2}$
$a\ne 0$ $\therefore a-d=4d$ or $a=5d$
$\therefore$ sides are $a-d,aa,a+d$
or $4d,5d,6d$
Hence the required ratio $4:5:6$.