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

Let $PM$ be the tower.

Given, $∠ BAP = \alpha, ∠ CBP = 2\alpha and ∠ MCP = 3\alpha$

$∴ ∠ APB = ∠ PBC - ∠ BAP = 2\alpha - \alpha = \alpha$

Similarly, $∠ BPC = \alpha$

$⇒ BP = AB$

By sine formula in triangle $BCP$, we have

$\dfrac{BC}{\sin ( ∠ BPC )} = \dfrac{BP}{\sin ( ∠ BCP )}$

$⇒ \dfrac{BC}{sin\alpha} = \dfrac{AB}{\sin ( {180}^{o} - 3\alpha )}$

$⇒ \dfrac{AB}{BC} = \dfrac{\sin 3\alpha}{sin\alpha} = \dfrac{3sin\alpha - 4\sin^{3}\alpha}{sin\alpha}$

$= 3 - 4\sin^{2}\alpha$

$= 3 - 2( 1 - \cos 2\alpha )$

$= 1 + 2\cos 2\alpha$