Question

A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60o. When he moves 40 metres away from the bank, he finds the angle of elevation to be 30o. The height of the tree and the width of the river are respectively.

Answer 1

Let AB be the tree of height h meters, C be a point at a distance of x meters from the base of the tree such that the angle of elevation of the top at C is 60° and D be the point at a distance of 40 meters from C such that the angle of elevation at D is 30°.

Now, In ΔCAB,

Tan 60° $= \frac{h}{x}$
⇒ $\sqrt{3} = \frac{h}{x}$
⇒  $h = \sqrt{3x}$ -----(i)

And, In ΔDAB,

Tan 30° $= \frac{h}{40 + x}$

⇒ $\frac{1}{ \sqrt{3} } = \frac{h}{40 + x}$
⇒ $40 + x = \sqrt{3h}$ -----(ii)

Substituting $h = \sqrt{3x}$  in equation (ii)

⇒ $40 + x = \sqrt{3x}\sqrt{3x}$
⇒ $40 + x = \sqrt{3x}$
⇒ $2x = 40$

∴ $x = 20$

Therefore, width of the river = 20 meters

Thus, $h = \sqrt{3x}$ 
             $= 20 \sqrt{3}$

∴ The height of the tree is 20√3 m.

Hope This Helps :)

Answer 2

$\textbf{Answer is in Attachment !}$