Question

The angle of elevation of the top of a tree from a point A on the ground is ${60}^{\circ }$ . On walking 20 m away from point A, to a point B, the angle of elevation changes to ${30}^{\circ }$. Find the height of the tree.

• A. $10\sqrt{3}$ m
• B. $20\sqrt{3}$ m
• C. $30\sqrt{3}$ m
• D. $40\sqrt{3}$ m

Answer 1

The correct option is A

$10\sqrt{3}$ m

In triangle ACD,

$\tan {60}^{\circ }=\frac{DC}{CA}$

$\Rightarrow \sqrt{3}=\frac{h}{x}$ ..................... (1)

In triangle CDB,

$\tan {30}^{\circ }=\frac{CD}{CB}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+20}$....................... (2)

Using (1) and (2), we get

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{x\sqrt{3}}{x+20}$

$\Rightarrow x+20=3x$

$\Rightarrow 2x=20$

$\Rightarrow x=10$

$\therefore h=x\sqrt{3}=10\sqrt{3}$

Thus, the height of the tree is $10\sqrt{3}m$.