Question

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.

✓Content quality answer needed​

Answer 1

AnswEr:

$\star\;{\underline{\sf{\purple{As\;per\;givEn\; Question:-}}}}$

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

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

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

$\;\bullet\:\sf\ \textless QYR = \bf\ 30^{\circ}$

$\;\bullet\:\sf\ \textless QXP = \bf\ 60^{\circ}$

Reference of image is in diagram

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

$\rule{170}2$

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

$\normalsize\bullet\:\sf\ Let \; PX = RY \; be \; \bf\ B$

☯ $\normalsize\sf In \; \triangle QYR$

$\dashrightarrow\sf Tan30^\circ = \dfrac{QR}{RY}$

$\small\sf\dag\ As \; we \; know, \; Tan30^{\circ} = \dfrac{1}{\sqrt{3}}$

$\normalsize\dashrightarrow\sf\dfrac{1}{\sqrt{3}}= \dfrac{54 - h}{B}$

$\small\sf{\dag\ Using \: cross \: multiplication\;-}$

$\normalsize\dashrightarrow\sf\ B = (54 - h)\sqrt{3}\qquad\bigg\lgroup\bf Eq.(1) \bigg\rgroup$

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

$\normalsize\dashrightarrow\sf\ Tan60^{\circ} = \dfrac{PQ}{PX}$

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

$\normalsize\dashrightarrow\sf\ \sqrt{3} = \dfrac{54}{B}$

$\small\sf{\dag\ Now, \; Using \: cross \: multiplication\;-}$

$\normalsize\dashrightarrow\sf\ B = (54 - h)\sqrt{3}\qquad\bigg\lgroup\bf Eq.(2) \bigg\rgroup$

$\rule{170}1$

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

$\normalsize\dashrightarrow\sf\ (54 - h)\sqrt{3} = \dfrac{54}{\sqrt{3}}$

$\normalsize\dashrightarrow\sf\ (54 - h) = \dfrac{54}{\sqrt{3} \times\ \sqrt{3}}$

$\normalsize\dashrightarrow\sf\ (54 - h) = \dfrac{54}{3}$

$\normalsize\dashrightarrow\sf\ h = \dfrac{162 - 54}{3}$

$\normalsize\dashrightarrow\sf\ h = \dfrac{\cancel{108}}{\cancel{3}}$

$\normalsize\dashrightarrow\sf\pink{h = 36m}\;\bigstar$

$\sf\therefore\; \underline{Hence,\;the\;height\;of\;other\;temple\;is\;36m}$

Answer 2

☢ Solution :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\linethickness{0.4mm}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){4.4}}\put(8,3){\line(3,0){3}}\put(11.1,2){\sf50 - h}\put(7.3,2){\sf50 - h}\put(11.1,4){\sf h}\qbezier(8,1)(10.7,5)(11,5.4)\qbezier(8,3)(10.5,5)(11,5.4)\qbezier(8.65,1)(8.7,1.4)(8.4,1.6)\put(8.7,1.2){\sf60^{\circ} }\qbezier(8.5,3.4)(8.8,3.2)(8.7,3)\put(8.8,3.2){\sf 30^{\circ} }\put(7.7,3){\large{A}}\put(7.7,1){\large{B}}\put(11.1,1){\large{C}}\put(11.1,3){\large{D}}\put(11.1,5.3){\large{E}}\put(9.5,0.8){\sf x}\put(9.5,2.8){\sf x}\end{picture}$

• Let AB and CE are two temples each at the bank of the river.

• Let CE = 50m and AB = H m and $\angle$CBE = 60° and $\angle$DAE = 30°

In ∆ADE,

$:\implies\sf tan\:30^{\circ} = \dfrac{ED}{AD}\\\\\\:\implies\sf tan\:30^{\circ} = \dfrac{50}{x}\\\\\\:\implies\sf \dfrac{1}{\sqrt {3}} = \dfrac{h}{x}\\\\\\:\implies\sf x = h{\sqrt{3}}$

In ∆BCE,

$:\implies\sf tan\:60^{\circ} = \dfrac{EC}{BC}\\\\\\:\implies\sf tan\:60^{\circ} = \dfrac{50}{x} \\\\\\:\implies\sf \sqrt {3} = \dfrac{50}{x}\\\\\\:\implies\sf 50 = \sqrt 3 \times xh \sqrt 3\\\\\\:\implies\sf h = \dfrac{50}{3}$

$\underline{\boldsymbol{ Now\: the\: distance\: between \: temples \: :}}$

$:\implies\sf x = h \sqrt 3 \\\\\\:\implies\sf x = \dfrac{50}{3} \times \sqrt 3\\\\\\:\implies\sf x = \dfrac{50}{3}$

☯ $\underline{\boldsymbol{Height\: of\: the\: temple \: :}}$

$:\implies\sf Height = 50 - \dfrac{50}{ 3 }\\\\\\:\implies\sf Height = \dfrac{150}{3} - \dfrac{150}{3}\\\\\\:\implies\sf Height = \dfrac{100}{3}\\\\\\:\implies\sf Height = 33.33\:m$

$\therefore\:\underline{\textsf{Height of the other temple is \textbf{33.33 m}}}.$