Question

The Angle of elevation of the top of a tower from a point on the ground is 60°. At a point, 40 m vertically above the point on the ground, the Angle of elevation is 45°. Find the Height of Tower.​

Answer 1

AnswEr :

⋆ Refrence of Image is in the Diagram :

$\setlength{\unitlength}{1cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\large{Y}}\put(7.6,1){\large{X}}\put(11.1,1){\large{P}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){4.4}}\put(8,3){\line(3,0){3}}\put(11.1,2){40 m}\put(7.2,2){40 m}\put(11.1,4){x m}\put(8,1){\line(2,3.){2.9}}\put(8,3){\line(5,4){3}}\put(11.1,3){\large{R}}\put(11.1,5.3){\large{Q}}\put(8.2,1.08){\circle{0.2}}\put(8.4,1.2){ 60^{\circ} }\put(8.1,3.03){\circle{0.15}}\put(8.4,3.1){ 45^{\circ} }\end{picture}$

• In ∆ YRQ, we have :

$\implies\tt\tan(45 \degree) = \dfrac{QR}{YR} \\ \\\implies \tt1 = \dfrac{x}{YR} \\ \\\implies \tt YR = x \\ \\\implies \tt XP = x \qquad \qquad [ \because YR = XP]$

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• In ∆ XPQ, we have :

$\implies\tt\tan(60 \degree) = \dfrac{PQ}{PX} \\ \\\implies\tt \sqrt{3} =\dfrac{(x + 40)}{x} \\ \\\implies\tt \sqrt{3}x = x + 40 \\ \\\implies\tt \sqrt{3}x - x =40 \\ \\\implies\tt x( \sqrt{3} - 1) = 40 \\ \\\implies\tt x = \dfrac{40}{( \sqrt{3} - 1)} \\\\\implies\tt x = \dfrac{40}{( \sqrt{3} - 1) } \times\dfrac{( \sqrt{3} + 1)}{(\sqrt{3} + 1)} \\ \\\implies\tt x = \dfrac{40( \sqrt{3} + 1)}{ {( \sqrt{3} )}^{2} - {(1)}^{2} } \\ \\\implies\tt x = \dfrac{40( \sqrt{3} + 1)}{3 - 1} \\ \\\implies\tt x = \dfrac{ \cancel40(\sqrt{3} + 1)}{\cancel2} \\ \\\implies\tt x =20( \sqrt{3} + 1) \\ \\\implies\tt x =20(1.732 + 1) \qquad \scriptsize[\sqrt{3} = 1.732]\\ \\\implies\tt x =20 \times 2.732 \\\\\implies \boxed{\pink{\tt x =54.64 \:metres}}$

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• H E I G H T⠀O F⠀T O W E R :

$\longrightarrow \tt Height \:of \:Tower = PQ\\\\\longrightarrow \tt Height \:of \:Tower = (QR+PR)\\\\\longrightarrow \tt Height \:of \:Tower = (x+40\:metres)\\\\\longrightarrow \tt Height \:of \:Tower = (54.64 + 40)\:metres\\\\\longrightarrow\large\boxed{\red{\tt Height \:of \:Tower = 94.64\:metres}}$

⠀

∴ Height of the tower is 94.64 metres.

Answer 2

Answer:

$\bold\red{Height\:of\:tower=94.64\:m}$

_____________________

Let OM = x meter.

So, height of tower = (x + 40) m

In ∆OPM,

$\tt tan(45) \degree = \frac{OM}{OP}$

$\rightarrow\tt 1= \frac{x}{OP}$

$\rightarrow\tt OP = x$

$\because OP = QN$

$\therefore QN = x$

________________________

In ∆MNQ,

$\tt tan(60)\degree =\frac{MN}{QN}$

$\rightarrow\tt \sqrt{3}=\frac{x+40}{x}$

$\rightarrow\tt \sqrt{3}x=x+40$

$\rightarrow\tt \sqrt{3}x - x = 40$

$\rightarrow\tt x(\sqrt{3}-1)=40$

$\rightarrow\tt x =\frac{40}{\sqrt{3}-1}$

$\rightarrow\tt x =\frac{40(\sqrt{3}+1)}{(\sqrt{3}-1)+(\sqrt{3}+1)}$

$\rightarrow\tt x = \frac{40(\sqrt{3}+1)}{ {(\sqrt{3}) }^{2}-1}$

$\rightarrow\tt x = \frac{40(\sqrt{3}+1)}{3-1}$

$\rightarrow\tt x = \frac{40(\sqrt{3}+1)}{2}$

$\rightarrow\tt x = 20(\sqrt{3}+1)$

$\rightarrow\tt x = 20(1.732+1)$

$\rightarrow\tt x = 54.64\:m$

_____________________

Height of tower = MN =(x + 40) m

$\mapsto$ Height of tower = (54.64+40) m

$\mapsto$ Height of tower = 94.64 m

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