Let a vertical tower AB have its end A on the level ground. Let C...

Let $∠ APC = \alpha$
Given, $AB = 2\left(AC\right) = 2\left(BC\right)$

Also $AP = 2AB ⇒ AP = 4\left(AC\right) = 4\left(BC\right)$

$⇒ tan\alpha = \dfrac{AC}{AP} = \dfrac{AC}{4\left(AC\right)} = \dfrac{1}{4}$

$⇒ \tan \left(\alpha + \beta\right) = \dfrac{AB}{AP} = \dfrac{AB}{2\left(AB\right)} = \dfrac{1}{2}$

Let, $tan\beta = x$

$\dfrac{tan\alpha + tan\beta}{1 - tan\alpha ⋅ tan\beta} = \dfrac{1}{2}$

$⇒ \dfrac{\dfrac{1}{4} + x}{1 - \dfrac{x}{4}} = \dfrac{1}{2}$

$⇒ \dfrac{1 + 4x}{4 - x} = \dfrac{1}{2}$

$⇒ 2 + 8x - 4 + x = 0$

$⇒ 9x - 2 = 0$

$⇒ x = \dfrac{2}{9}$