Two poles of equal heights are standing opposite to each other on either side of the road,
which is 120 feet wide. From a point between them on the road, the angles of elevation
of the top of the poles are 60° and 30° respectively. Find the height of the poles and the
distances of the point from the poles.
Answers
Step-by-step explanation:
Letheightoftheequalpoles
AB=CD=hfeet
Distance \: between \:two \:poles (BC) = 120\ftDistancebetweentwopoles(BC)=120\ft
Point \:of \: observation \: is \: EPointofobservationisE
Let \: BE = x \:ftLetBE=xft
EC = ( 120 - x ) \:ftEC=(120−x)ft
\begin{gathered}i ) In \: triangle ABE , \\tan \angle {AEB} = \frac{AB}{BE}\end{gathered}
i)IntriangleABE,
tan∠AEB=
BE
AB
\implies tan 60\degree = \frac{h}{x}⟹tan60°=
x
h
\implies \sqrt{3} x = h \: ---(1)⟹
3
x=h−−−(1)
\begin{gathered}ii )In \: triangle DEC , \\tan \angle {DEC} = \frac{DC}{EC}\end{gathered}
ii)IntriangleDEC,
tan∠DEC=
EC
DC
\implies tan 30\degree = \frac{h}{(120-x)}⟹tan30°=
(120−x)
h
\implies \frac{1}{\sqrt{3} } (120 - x ) = h \: ---(2)⟹
3
1
(120−x)=h−−−(2)
/* From (1) and (2) */
\sqrt{3} x = \frac{1}{\sqrt{3} } (120 - x )
3
x=
3
1
(120−x)
\implies \sqrt{3}^{2} x = 120 - x⟹
3
2
x=120−x
\implies 3x = 120 - x⟹3x=120−x
\implies 4x = 120⟹4x=120
\begin{gathered}\implies x = \frac{120}{4} \\= 30\:ft\: --(3)\end{gathered}
⟹x=
4
120
=30ft−−(3)
/* put value of x in equation (1) , we get */
h = \sqrt{3} \times 30h=
3
×30
\implies h =30 \sqrt{3}⟹h=30
3
\implies h =39 \times 1.732⟹h=39×1.732
\implies h = 51.96 \:ft⟹h=51.96ft
Therefore.,
\red{ Height \:two \:equal \:poles } \green {= 51.96\:ft }Heighttwoequalpoles=51.96ft
\red{ Distance \:from \:B \: to \:the \:point} \green {= 30 \:ft }DistancefromBtothepoint=30ft
\begin{gathered}\red{ Distance \:from \:C \: to \:the \:point} \\=(120 - x)\\= (120-30) \\\green {= 90 \:ft }\end{gathered}
DistancefromCtothepoint
=(120−x)
=(120−30)
=90ft
