Question

Applications of Trigonometry
301
ple-, Two men on either side of a femple of 30 meter height observe its top at the anges
evation 30 and 60' respectively, Find the distance between the two men,​

Answer 1

Given :-

• Height of the temple = 30 m.
• Angle of elevation of the top of temple observed from a point (say, B) = 60°
• Angle of elevation of the top of temple observed from another point (say, C) on the other side = 30°

To Find :-

• Distance between the two men (BC).

Solution :-

Figure:-

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){69.282}}\put(0,0){\line(2,3){17.32}}\put(69.282,0){\line(-2,1){51.96}}\put(17.32,0){\line(0,1){26}}\qbezier(2,0)(3,0)(2,3)\qbezier(63,0)(63,1)(64,2.5)\footnotesize\put(3.5,1.5){ \sf{60^o} }\put(56,1.3){ \sf{30^o} }\put(16,28){\sf{A}}\put(-3,-4){\sf{B}}\put(16.5,-4){\sf{D}}\put(69.282,-4){\sf{C}}\put(19,11){\sf{30\ m}}\put(7.5,-4){\sf{x}}\put(42,-4){\sf{y}}\put(6,-3){\vector(-1,0){6}}\put(10.5,-3){\vector(1,0){5.5}}\put(41,-3){\vector(-1,0){21}}\put(45,-3){\vector(1,0){23.5}}\end{picture}$

In fig, AD represents the temple of height 30 m. B & D are the two points from where two men observed the angle of elevation of the top of temple as 60° and 30° respectively. x represents BD and y represents DC. BC (x + y) is the distance between the two men.

$\hrulefill$

In △ ABD,

$\implies\rm{tan\:60^{\circ}=\dfrac{AD}{BD}=\dfrac{30}{x}}$

$\implies\rm{\sqrt{3}=\dfrac{30}{x}}$

$\implies\rm{x=\dfrac{30}{\sqrt{3}}=10\sqrt{3}\:m\longrightarrow{(1)}}$

In △ ACD,

$\implies\rm{tan\:30^{\circ}=\dfrac{AD}{DC}=\dfrac{30}{y}}$

$\implies\rm{\dfrac{1}{\sqrt{3}}=\dfrac{30}{y}}$

$\implies\rm{y=30\sqrt{3}\:m\longrightarrow{(2)}}$

Adding ( 1 ) and ( 2 ), we get,

$\implies\rm{x+y=10\sqrt{3}+30\sqrt{3}}$

$\implies\rm{BC=}\:\bold{40\sqrt{3}\:m.}$

Hence,

• Distance between the two men (BC) = 40√3 m.