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Question
The Horizontal distance between two towers is 140m. The angle of elevation of the top of the first tower, when seen from the top of the second tower is 30 degree.If the height of the second tower is 60m,find the height of the first tower.
The Horizontal distance between two towers is 140m. The angle of elevation of the top of the first tower, when seen from the top of the second tower is 30 degree.If the height of the second tower is 60m,find the height of the first tower.
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Solution
Let AB be the height of second tower and CD be the height of first tower.
Given, BD = AE = 140 m
& AB = DE = 60 m
In ΔAEC,
tan 30°= Perpendicular /Base= CE/AE
tan 30 ° = CE/140
1/√3= CE /140
CE= 140 / √3
CE= 140 ×√3/ (√3×√3)
[ Rationalising the denominator]
CE= 140√3 /3
Height of the first tower CD= CE+DE
= 140√3/3 + 60
=( 140 × 1.73) /3 + 60
[ √3=1.73]
= 242.2/3 + 60
= 80.73 + 60
= 140.73 m
Height of the first tower (CD)=140.73 m
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Given, BD = AE = 140 m
& AB = DE = 60 m
In ΔAEC,
tan 30°= Perpendicular /Base= CE/AE
tan 30 ° = CE/140
1/√3= CE /140
CE= 140 / √3
CE= 140 ×√3/ (√3×√3)
[ Rationalising the denominator]
CE= 140√3 /3
Height of the first tower CD= CE+DE
= 140√3/3 + 60
=( 140 × 1.73) /3 + 60
[ √3=1.73]
= 242.2/3 + 60
= 80.73 + 60
= 140.73 m
Height of the first tower (CD)=140.73 m
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