A person standing at the foot of a tower walks a distance of 3...

Let $O$ be the foot of the tower $OP$

$OA = 3m, AB = 4m$
$∴ OB = 5m\left\{∵ OA^{2} + AB^{2} = OB^{2}\right\}$
$∴$ In $∆ AOP,$
$\tan 30 ° = \dfrac{h}{3} ∴ h = \sqrt{3}$
and in $∆ OPB,$

$tan \beta = \dfrac{h}{5} ∴ h = 5tan \beta$
Thus, $h = \sqrt{3} = 5tan \beta ⇒ tan \beta = \dfrac{\sqrt{3}}{5}$
$\cos 2\beta = \dfrac{1 - \tan^{2} \beta}{1 + \tan^{2} \beta} = \dfrac{1 - \dfrac{3}{25}}{1 + \dfrac{3}{25}} = \dfrac{11}{14}$