Question

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​

Answer 1

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

Answer 2

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