Q186 / 270 · NDA 2017 (Phase 2) · Mathematics
The angles of elevation of the top of a tower from the top and foot of a pole are respectively $30^{\circ}$ and $45^{\circ}$. If $\mathrm{h}_{\mathrm{T}}$ is the height of the tower and $h_p$ is the height of the pole, then which of the following are correct?
1. $\frac{2 \mathrm{~h}_{\mathrm{P}} \mathrm{h}_{\mathrm{T}}}{3+\sqrt{3}}=\mathrm{h}_{\mathrm{P}}^2$
2. $\frac{\mathrm{h}_{\mathrm{T}}-\mathrm{h}_{\mathrm{P}}}{\sqrt{3}+1}=\frac{\mathrm{h}_{\mathrm{P}}}{2}$
3. $\frac{2\left(\mathrm{~h}_{\mathrm{P}}+\mathrm{h}_{\mathrm{T}}\right)}{\mathrm{h}_{\mathrm{P}}}=4+\sqrt{3}$
Select the correct answer using the code given below.
- A$1$ and $3$ only
- B$2$ and $3$ only
- Correct answerC$1$ and $2$ only
- D$1,2$ and $3$
Solution
Let the distance between tower and pole be ' $x$ '.

$\begin{aligned} & \tan 45^{\circ}=\frac{\mathrm{h}_{\mathrm{t}}}{\mathrm{x}} \Rightarrow \frac{\mathrm{h}_{\mathrm{t}}}{\mathrm{x}}=1 \\ & \Rightarrow \mathrm{~h}_{\mathrm{t}}=\mathrm{x} .\end{aligned}$ Now, $\tan 30^{\circ}=\frac{h_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{x}} \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{x}}$ $\begin{aligned} & \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{h}_{\mathrm{t}}} \\ & \Rightarrow \frac{1}{\sqrt{3}-1}=\frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{h}_{\mathrm{t}}-\left(\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}\right)}\end{aligned}$ $\Rightarrow \frac{\sqrt{3}+1}{2}=\frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{h}_{\mathrm{p}}} \Rightarrow \frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\sqrt{3}+1}=\frac{\mathrm{h}_{\mathrm{p}}}{2}$\ldots(1) So, statement ' 2 ' is correct. Now, $\frac{h_t-h_p}{h_p}=\frac{\sqrt{3}+1}{2} \Rightarrow \frac{h_t-h_p+2 h_p}{h_p}=\frac{\sqrt{3}+1+4}{2}$ $\Rightarrow \frac{\mathrm{h}_{\mathrm{t}}+\mathrm{h}_{\mathrm{p}}}{\mathrm{h}_{\mathrm{p}}}=\frac{5+\sqrt{3}}{2}$\ldots(2) So, statement ‘3’ is incorrect.

$\begin{aligned} & \tan 45^{\circ}=\frac{\mathrm{h}_{\mathrm{t}}}{\mathrm{x}} \Rightarrow \frac{\mathrm{h}_{\mathrm{t}}}{\mathrm{x}}=1 \\ & \Rightarrow \mathrm{~h}_{\mathrm{t}}=\mathrm{x} .\end{aligned}$ Now, $\tan 30^{\circ}=\frac{h_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{x}} \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{x}}$ $\begin{aligned} & \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{h}_{\mathrm{t}}} \\ & \Rightarrow \frac{1}{\sqrt{3}-1}=\frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{h}_{\mathrm{t}}-\left(\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}\right)}\end{aligned}$ $\Rightarrow \frac{\sqrt{3}+1}{2}=\frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\mathrm{h}_{\mathrm{p}}} \Rightarrow \frac{\mathrm{h}_{\mathrm{t}}-\mathrm{h}_{\mathrm{p}}}{\sqrt{3}+1}=\frac{\mathrm{h}_{\mathrm{p}}}{2}$\ldots(1) So, statement ' 2 ' is correct. Now, $\frac{h_t-h_p}{h_p}=\frac{\sqrt{3}+1}{2} \Rightarrow \frac{h_t-h_p+2 h_p}{h_p}=\frac{\sqrt{3}+1+4}{2}$ $\Rightarrow \frac{\mathrm{h}_{\mathrm{t}}+\mathrm{h}_{\mathrm{p}}}{\mathrm{h}_{\mathrm{p}}}=\frac{5+\sqrt{3}}{2}$\ldots(2) So, statement ‘3’ is incorrect.
Hence, the correct answer is option: 'C'.