Question

$ABCD$ 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 $\theta$, then $\tan \theta$ is equal to

• A. $\sqrt{6}$
• B. $\frac{1}{\sqrt{6}}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{\sqrt{2}}{2}$

Answer 1

The correct option is B

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

Explanation for the correct option:

Step-1: Draw the figure and mark all angles:

Let side of suare is $a$

From the figure,

In $\Delta PCD$, we have

$\tan 30°=\frac{h}{a}$

$\Rightarrow$ $\frac{h}{a}=\frac{1}{\surd 3}$ …… $\left(1\right)$ $\left[\because \tan 30°=\frac{1}{\sqrt{3}}\right]$

In $\Delta ABD$, we have

$\begin{array}{rcl}BD& =& {\sqrt{(A{B}^2+AD)}}^2\\ & =& a\sqrt{2}\end{array}$

Now, in $\Delta PBD$

$PD=h$

Now, $\begin{array}{rcl}\tan \theta & =& \frac{PD}{BD}\end{array}$

Step- 2: Substitute the values of $PD$ and $BD$:

$\Rightarrow$$\tan \theta =\frac{h}{a\sqrt{2}}$

$\Rightarrow$$\tan \theta =\frac{a}{\sqrt{3}}\times \frac{1}{a\sqrt{2}}$

$\Rightarrow$$\tan \theta =\frac{1}{\sqrt{2}}x\frac{1}{\sqrt{3}}$

$\Rightarrow$$\tan \theta =\frac{1}{\sqrt{6}}$

Hence, option ‘B' is correct.