A tower subtends an angle 75 ^ o at a point on the same level as...

From the figure, in right-angled $∆ ABD,$
$\dfrac{AB}{BD} = tan {75}^{o}$
$∴ \dfrac{h}{x} = tan {75}^{o} ⇒ h = xtan {75}^{o} \ldots \ldots . . \left(i\right)$
In right-angled $∆ BCE,$

$\dfrac{BE}{EC} =$ tan {15}^{o}$<br />$⇒ $\dfrac{10}{x} =$ tan {15}^{o} ∴ x = $\dfrac{10}{tan {15}^{o}} \ldots \ldots\left(ii\right)$
From \left(i\right)$and \left(ii\right),$ we get,
$∴ h = \dfrac{10}{tan {15}^{o}} \times tan {75}^{o}$

$= 10\left(2 + \sqrt{3}\right)2$& meters