The angle of elevation of a tower from a point in the horizontal to the foot of the tower is 60º, if the distance of point from the foot of tower is 100 metres, the height of the tower is
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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
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