the angle of elevation of a tower is observed to be 600 at end of original base of 100 metres measured from its feet find the height of the tower
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Step-by-step explanation:
Answer
Let the height of the tower AC be h metres Given distance BC = 100 m, ∠ABC=60
∘
∴tan60
∘
=
BC
AC
=
100
h
⇒h=100×tan60
∘
=(100×
3
)metre
=100×1.732=173.2m
solution
Answered by
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Answer:
Let the height of the tower AC be h metres Given distance BC = 100 m, ABC = 60° =
.. tan 60°
tan 60°
AC h
BC 100
(100 x v3)metre
V
= 100 x 1.732 = 173.2 m
= 100 xtan 60°
AC h
=
BC 100
= (100 x v3)metre V
= 100 x 1.732 173.2 m
= 100 x
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h Tower
60°
B
Eye
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100 m
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