ABC is a triangular park with AB = AC = 100 metres. A vertical...

Let $AD = y \text{ and } DE = h$ (Tower)

$∴ {y}^{2} = {100}^{2} - {x}^{2}$

Now $cot\theta = 3\sqrt{2}$

$⇒ tan\theta = \dfrac{1}{3\sqrt{2}} = \dfrac{h}{y}$

$⇒ y = 3\sqrt{2}h$

$⇒ {y}^{2} = {100}^{2} - {x}^{2} = 18{h}^{2} \ldots$\left(1\right)$</p><p>$⇒ tan∅ = \dfrac{1}{\sqrt{7}} = \dfrac{h}{x}$</p><p>$⇒ x = h\sqrt{7} \ldots \left(2\right)$</p><p>Form \left(1\right)$ & \left(2\right)${100}^{2} - 7{h}^{2} = 18{h}^{2}$

$⇒ {25h}^{2} = {100}^{2} ⇒ h = 20$