Question

24. There are two temples, one on each bank of a river.
just opposite to each other. One temple is 54 m high.
From the top of this temple, the angles of depression
of the top and the foot of the other temple are 30
and 60° respectively. Find the height of the other
temple.​

Answer 1

$\bf{\underline{\underline{Given\: :}}}$

$\mathsf{Let, \: AB\: = \:54\: m }$

$\mathsf{Angle\: of \:depression\: of \:C \:and\: D}$

$\mathsf{from\:A\: be \:30°\: and \:60°\: respectively}$

$\mathsf{Let\: BD\: =\: x\: m}$

$\bf{\underline{\underline{To\:find\: :}}}$

$\mathtt{\diamond\:\:tan\:60° = \frac{AB}{BD}}$

$\mathtt{\rightarrow\:tan\:60° = \frac{54}{x}}$

$\mathtt{\rightarrow\: \sqrt{3} = \frac{54}{x}}$

$\mathtt{\rightarrow\: x = \frac{54}{\sqrt{3}}$

$\mathtt{ \rightarrow \: x = \Large{\frac{\sqrt{3} × \sqrt{3} × 18}{\sqrt{3}}}\rightarrow \: (1)}</p><p></p><p>[tex]\mathtt{\diamond\:\:tan\:30° = \frac{AB - CD}{BD}}$

$\mathtt{\rightarrow\:\:tan\:30° = \frac{54 - CD}{x}}$

$\mathtt{\rightarrow\: \frac{1}{\sqrt{3}} = \frac{54 - CD}{x}}$

$\mathtt{\rightarrow\: x = \sqrt{3}(54 - CD)}$

$\mathtt{\rightarrow\: x = 54\sqrt{3} - \sqrt{3} \:CD}$

$\mathtt{\rightarrow\: \sqrt{3}\:CD = 54\sqrt{3} - x}$

$\mathtt{\rightarrow\: \sqrt{3}\:CD = 54\sqrt{3} - 18\sqrt{3}\: \: (from\: 1)}$

$\mathtt{\rightarrow\: \sqrt{3}\:CD = 36\sqrt{3}}$

$\mathtt{\rightarrow\: CD = \frac{36 \: \cancel{\sqrt{3}}} \cancel{{\sqrt{3}}}}$

$\Large{\fbox{\mathtt{\green{\rightarrow\: CD = 36\: m}}}}$

$\mathtt{Therefore,\: height\: of\: the}$$\mathtt{other\: temple\: = 36\: m}$

Answer 2

AnswEr :

⋆ Reference of Image is in Diagram

$\setlength{\unitlength}{1cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\large{Y}}\put(7.6,1){\large{X}}\put(11.1,1){\large{P}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){4.4}}\put(8,3){\line(3,0){3}}\put(11.1,2){h}\put(7.2,2){h}\put(11.1,4){54 - h}\put(8,1){\line(2,3.){2.9}}\put(8,3){\line(5,4){3}}\put(11.1,3){\large{R}}\put(11.1,5.3){\large{Q}}\put(8.2,1.08){\circle{0.2}}\put(8.4,1.2){ 60^{\circ} }\put(8.1,3.03){\circle{0.15}}\put(8.4,3.1){ 30^{\circ} }\end{picture}$

$\normalsize\bullet\:\bf\ PQ \: and \: XY \: \sf\ are \: two \: temples$

$\normalsize\bullet\:\bf\ PX \: \sf\ is \: river$

$\normalsize\bullet\:\sf\ PQ = \bf\ 54m$

$\normalsize\bullet\:\sf\ \textless QYR = \bf\ 30^{\circ} \: \sf\ \textless QXP = \bf\ 60^{\circ}$

$\rule{170}2$

$\underline{\bigstar\:\textsf{Let's \: head \: to \: the \: question \: now:}}$

$\normalsize\bullet\:\sf\ Let \: PX \: be \: \sf\ B$

$\normalsize\bullet\:\sf\ PX = RY = \bf\ B$

☯ $\normalsize\sf\ In \: \triangle QYR :-$

$\normalsize\ : \implies\quad\sf\ Tan30^{\circ} = \frac{QR}{RY}$

$\scriptsize\sf{\qquad\dag\ Tan30^{\circ} = \frac{1}{\sqrt{3}} }$

$\normalsize\ : \implies\quad\sf\frac{1}{\sqrt{3}}= \frac{54 - h}{B}$

$\scriptsize\sf{\qquad\dag\ Using \: cross \: multiplication}$

$\normalsize\ : \implies\quad\sf\ B = (54 - h)\sqrt{3} \: --- (Eq.1)$

☯ $\normalsize\sf\ In \: \triangle QXP :-$

$\normalsize\ : \implies\quad\sf\ Tan60^{\circ} = \frac{PQ}{PX}$

$\scriptsize\sf{\qquad\dag\ Tan60^{\circ} = \sqrt{3} }$

$\normalsize\ : \implies\quad\sf\ \sqrt{3} = \frac{54}{B}$

$\scriptsize\sf{\qquad\dag\ Using \: cross \: multiplication}$

$\normalsize\ : \implies\quad\sf\ B = \frac{54}{\sqrt{3}} \: ---(Eq.2)$

$\rule{170}1$

☯ $\normalsize\sf\ Substitute \: the \: value \: of \: B \: from \: 1 \: \& \: 2 :-$

$\normalsize\twoheadrightarrow\sf\ (54 - h)\sqrt{3} = \frac{54}{\sqrt{3}}$

$\normalsize\twoheadrightarrow\sf\ (54 - h) = \frac{54}{\sqrt{3} \times\ \sqrt{3}}$

$\normalsize\twoheadrightarrow\sf\ (54 - h) = \frac{54}{3}$

$\normalsize\twoheadrightarrow\sf\ 54 - \frac{54}{3} = h$

$\normalsize\twoheadrightarrow\sf\ h = \frac{162 - 54}{3}$

$\normalsize\twoheadrightarrow\sf\ h = \frac{\cancel{108}}{\cancel{3}}$

$\normalsize\twoheadrightarrow\sf\red{h = 36m}$

$\therefore\:\underline{\textsf{Hence, \: the \: height \: of \: other \: temple \: is }{\textbf{\: 36m}}}$