The angle of elevation of the top of a tree from a point A on the ground is 60�

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.

Options:

A . 10√3 m

B . 20√3 m

C . 30√3 m

D . 40√3 m

Answer: Option A
:
A

In triangle ACD,

tan θ = \frac{D C}{C A} = \frac{h}{x}

\Rightarrow tan 60^{\circ} = \sqrt{3} = \frac{h}{x}

..................... 1
In triangle CDB,

\Rightarrow tan 30^{\circ} = \frac{C D}{C B}

\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 = 3 x

\Rightarrow 2 x = 20

\Rightarrow x = 10

∴ h = x \sqrt{3} = 10 \sqrt{3}

Thus, the height of the tree is

10 \sqrt{3} m

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