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$ABCD$ is a square plot. The angle of elevation of the top of a pole standing at $D$ from $A$ or $C$ is $30 °$ and that from $B$ is $θ$ , then $tanθ$ is equal to

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\sqrt{6}

\frac{1}{\sqrt{6}}

\frac{1}{\sqrt{3}}

\frac{2}{\sqrt{6}}

answer is B.

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Detailed Solution

It is given that

ABCD

is a square plot. The angle of elevation of the top of a pole which is standing at

D

from

A

C

30^{\circ}

and that from

B

θ

.
According to the given data the figure has been drawn below.
Question Image

Let the height of the pole be

h

m , then in

ΔPAD,

tan θ = P e r p e n d i c u l a r b a s e

\tan 30 ° = \frac{PD}{AD}

\Rightarrow \frac{h}{a} = \tan 30 °

\Rightarrow h = a tan 30 °

h = \frac{a}{\sqrt{3}}

……(1)

∵ tan 30 ° = 13

ΔPBD,

we have,

tan θ = P e r p e n d i c u l a r b a s e

tanθ = \frac{PD}{BD}

\Rightarrow \frac{h}{BD} = tanθ

h = BD \times tanθ

\Rightarrow h = \sqrt{(a^{2} + a^{2})} tanθ

\Rightarrow h = \sqrt{2 a^{2}} tanθ

\Rightarrow h = a \sqrt{2} tanθ

-------(2)
From

△ PBD,

diagonal

= a \sqrt{2}

.
From eq. 1 and 2,

\Rightarrow \frac{a}{\sqrt{3}} = a \sqrt{2} tanθ

\Rightarrow \frac{1}{\sqrt{3}} = \sqrt{2} tanθ

\Rightarrow tanθ = \frac{1}{\sqrt{2} \times \sqrt{3}}

\Rightarrow tanθ = \frac{1}{\sqrt{6}}

So,

tanθ = \frac{1}{\sqrt{6}}

.
So, the correct option is 2.

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