Question

Angles of elevation of the top of a tower from three points (collinear) $A,B,C$ on a road leading to the foot of the tower are $30°,45°,60°$. the ratio of $AB$ to $AC$

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

Answer 1

The correct option is A

$\sqrt{3}:1$

Find the ratio of $AB$ to $AC$:

According to the given information the figure should be as follows.

Let the height of tower$=h$

In $\begin{array}{rcl}& & ∆EDA,\end{array}$

$\begin{array}{rcl}\tan \left({30}^{°}\right)& =& \frac{\mathrm{ED}}{\mathrm{AD}}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{1}{\sqrt{3}}& =& \frac{\mathrm{ED}}{\mathrm{AD}}\end{array}$

$\begin{array}{rcl}& =& \frac{h}{AD},\end{array}$

$\Rightarrow$$\begin{array}{rcl}\tan \left({45}^{°}\right)& =& \frac{\mathrm{h}}{\mathrm{BD}}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}BD& =& h\end{array}$

In $\begin{array}{rcl}& & ∆EDC,\end{array}$

$\begin{array}{rcl}\tan \left({60}^{°}\right)& =& \frac{\mathrm{h}}{\mathrm{CD}}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}CD& =& \frac{\mathrm{h}}{\sqrt{3}}\end{array}$

Now,

$\begin{array}{rcl}\frac{\mathrm{AB}}{\mathrm{BC}}& =& \frac{\mathrm{AD}-\mathrm{BD}}{\mathrm{BD}-\mathrm{CD}}\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{\mathrm{AB}}{\mathrm{BC}}& =& \frac{\mathrm{h}\sqrt{3}-\mathrm{h}}{\mathrm{h}-{\displaystyle \frac{\mathrm{h}}{\sqrt{3}}}}\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{\mathrm{AB}}{\mathrm{BC}}& =& \frac{\sqrt{3}-1}{(\sqrt{3}-1)}\times \sqrt{3}\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{\mathrm{AB}}{\mathrm{BC}}& =& \frac{\sqrt{3}}{1}\end{array}$

$\begin{array}{rcl}\mathbf{\therefore }\mathit{A}\mathit{B}\mathbf{:}\mathit{B}\mathit{C}& \mathbf{=}& \sqrt{\mathbf{3}}\end{array}$

Hence, option $\left(A\right)$ is correct.