Question

The angle of elevation of a cloud $C$ from a point $P,200m$ above a still lake is ${30}^{°}$. If the angle of depression of the image of $C$ in the lake from the point $P$ is $60°$, then $PC$ (in m) is equal to

• A. $200\sqrt{3}$
• B. $400\sqrt{3}$
• C. $400$
• D. $100$

Answer 1

The correct option is C

$400$

Explanation for the correct option:

Find $PC$:

Using a given data to draw a figure:

Let $PA=x$

$$\begin{gathered} \tan {30}^{°}=\frac{AC}{PA} \\ \Rightarrow \frac{x}{\sqrt{3}}=AC \end{gathered}$$

$\begin{array}{rcl}A{C}^{\text{'}}& =& AB+B{C}^{\text{'}}\\ A{C}^{\text{'}}& =& AB+BC\left[\because BC\text{'}=BC\right]\\ A{C}^{\text{'}}& =& 200+200+\frac{x}{\sqrt{3}}\left[\because BC=200+\frac{x}{\sqrt{3}}\right]\\ A{C}^{\text{'}}& =& 400+\frac{x}{\sqrt{3}}\end{array}$

Now, In $∆PA{C}^{\text{'}}$

$$\begin{gathered} \tan {60}^{°}=\frac{A{C}^{\text{'}}}{PA} \\ \Rightarrow PA\sqrt{3}=A{C}^{\text{'}} \end{gathered}$$

Therefore,

$$\begin{gathered} x\sqrt{3}=400+\frac{x}{\sqrt{3}} \\ \Rightarrow x=200\sqrt{3} \end{gathered}$$

In $∆PAC$

$$\begin{gathered} \sin {30}^{°}=\frac{AC}{PC} \\ \Rightarrow \frac{1}{2}=\frac{{\displaystyle \frac{200\sqrt{3}}{\sqrt{3}}}}{PC}\left[\because AC=x\sqrt{3};x=200\sqrt{3}\right] \\ \Rightarrow PC=400m \end{gathered}$$

Hence, option (B) is the correct answer.