Question

A man is standing at a distance of 200m from Qutub Minar. He starts walking towards it and walks for 125m. The angle of elevation from the initial point becomes double as he reached there. Find the height of Qutub Minar.

Answer is 100m but I need proper solution.

Answer 1

❏ $\huge{\underline\mathrm{Question}}$

@ A man is standing at a distance of 200m from Qutub Minar. He starts walking towards it and walks for 125m. The angle of elevation from the initial point becomes double as he reached there. Find the height of Qutub Minar.

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❏ $\huge{\underline\mathrm{Answer}}$

$\sf\implies \text{Height of Q.M.= 100 m}$

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❏ $\huge{\underline\mathrm{Figure-}}$

$\setlength{\unitlength}{0.9 cm}\begin{picture}(12,4)\thicklines\put(5.6,9.1){ A }\put(5.6,5.8){ B }\put(12.2,5.9){ C }\put(8.8,5.6){ D }\put(11,6.1){ \alpha }\put(7.3,6.15){ \beta=2\alpha }\put(5.5,7.5){ H }\put(6,6){\line(1,0){6}}\put(6,9){\line(1,-1){3}}\put(6,9){\line(2,-1){6}}\put(6,6){\line(0,1){3}}\put(6,5){\line(1,0){2}}\put(10,5){\line(1,0){2}}\put(8.6,5){ 200\:m }\put(10,5.6){ 125\:m }\end{picture}$

❏ $\huge{\underline\mathrm{Given}}$

$\begin{cases}\text{Distance Between Q.M. and man = 200 m.} \\ \text{Covered Distance by walking = 125 m} \\ \text{Initial A.E. is doubled than Final A.E.}\end{cases}$

where,

• A.E. = angle Of Elevation.

• Q.M. = Qutub Minar.

❏ $\huge{\underline\mathrm{Solution}}$

Let, the initial angle of elevation = $\alpha$

and the final angle of elevation = $\beta$

and height of the Q.M. = H metres .

According to the Question final angle of elevation becomes doubles than the initial,

So, $\boxed{\beta\:=\:2\alpha}$

Case : 1

$\sf\implies \tan\alpha = \frac{AB}{BC}$

$\sf\implies \tan\alpha = \frac{H}{200}$

Case : 2

$\sf\implies \tan\beta = \frac{AB}{BD}$

$\sf\implies \tan2\alpha= \frac{H}{BC-CD}$

$\sf\implies \dfrac{2\tan\alpha}{1-\tan^2\alpha} = \dfrac{H}{200-125}$

$\sf\implies \dfrac{\cancel2\times(\dfrac{\cancel H}{\cancel200}}{1-(\dfrac{H}{200})^2} = \dfrac{\cancel H}{75}$

$\sf\implies \dfrac{1}{100\times(1-\frac{H^2}{40,000})} = \dfrac{1}{75}$

$\sf\implies \dfrac{1}{\times(1-\frac{H^2}{40,000})} = \dfrac{\cancel{100}}{\cancel{75}}$

$\sf\implies \dfrac{1}{\times(1-\frac{H^2}{40,000})} = \dfrac{4}{3}$

$\sf\implies 4\times(1-\frac{H^2}{40,000}) = 3$

$\sf\implies 4\:-\:\cancel4\times\frac{H^2}{\cancel{40,000}}= 3$

$\sf\implies \frac{H^2}{10,000}=4- 3$

$\sf\implies H^2=10,000$

$\sf\implies H=\sqrt{10,000}$

$\sf\implies \boxed{\large{\red{H=100\:m}}}$ (taking +ve term).

$\therefore$ Height of the Qutub Minar is =100 metres.

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❏ $\huge{\underline\mathrm{Formula Used}}$

➾$\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$

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