Question

A man standing on the bank of a river observes that the angle of elevation of the top of a tree just on the opposite bank is ${60}^{\circ }$. If the angle of elevation is ${30}^{\circ }$ from a point at a distance 'y' metres from the bank, then the height of the tree = ___ m.

• A. 30
• B. 20√3
• C. 20
• D. $2\sqrt{3}y$

Answer 1

The correct option is A

30

Let AB be the tree of height = h m
CA be the length of river = xm
CD = y

In $\Delta ABC$

$tan{60}^{\circ }=\frac{h}{x}$

$\sqrt{3}=\frac{h}{x}$

$x=\frac{h}{\sqrt{3}}\dots \left(i\right)$

In $\Delta ABD$

$tan{30}^{\circ }=\frac{h}{y+x}$

$\frac{1}{\sqrt{3}}=\frac{h}{y+\frac{h}{\sqrt{3}}}$

$\frac{1}{\sqrt{3}}=\frac{h}{\frac{y\sqrt{3}+h}{\sqrt{3}}}$

$\frac{1}{\sqrt{3}}=\frac{\sqrt{3}h}{y\sqrt{3}+h}$

$y\sqrt{3}+h=3h$

$2h=y\sqrt{3}$

$h=\frac{\sqrt{3}}{2}y$