JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Properties of Triangle

Let the smallest angle be $\theta$ and the largest, opposite the longest side, be $2\theta$; the third angle is then $\pi -3\theta$. With the sides in A.P., $2b=a+c$, and the sine rule turns this into $2\sin (\pi -3\theta )=\sin \theta +\sin 2\theta$, that is $2\sin 3\theta =\sin \theta +\sin 2\theta$. Expanding and simplifying leads to $8{\cos }^2\theta -2\cos \theta -3=0$, whose valid root is $\cos \theta =\frac{3}{4}$. The sides are proportional to the sines of their opposite angles, which work out to $1:\frac{5}{4}:\frac{3}{2}$, or $4:5:6$.