A tower subtends angles α, 2α and& 3α respectively at points, A, B...

Let, the height of the tower is $PQ = h$
Where, $Q$ is the foot of the tower and $P$ is the top of the tower, then
$QA = hcot \alpha$
$QB = hcot 2\alpha$
$QC = hcot 3\alpha$

$\dfrac{AB}{BC} = \dfrac{QA - QB}{QB - QC} = \dfrac{cot \alpha - \cot 2\alpha}{\cot 2\alpha - \cot 3\alpha}$
$= \dfrac{\dfrac{cos \alpha}{sin \alpha} - \dfrac{\cos 2\alpha}{\sin 2\alpha}}{\dfrac{\cos 2\alpha}{\sin 2\alpha} - \dfrac{\cos 3\alpha}{\sin 3\alpha}} = \dfrac{\sin 2\alpha cos \alpha - \cos 2\alpha sin \alpha}{\sin 3\alpha cos 2\alpha - \cos 3\alpha sin 2\alpha} \times \dfrac{\sin 3\alpha}{sin \alpha}$
$= \dfrac{\sin 3\alpha}{sin \alpha} = 3 - 4sin^{2}\alpha = 1 + 2\cos 2\alpha$