TS ICET 23 May 2019 Shift 2 Paper

Question : 121 of 200

Marks: +1, -0

The angle of elevation of a tower from a point 40 meters from foot of the tower is

cot^{-1}(3/5)

. Then the height (in meters) of the tower is

Solution:

Given

\cotθ=3/5

{3}/{5}={40}/{h}

h=\;{200}/{3}

=66{2}/{3} m

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