sir this is the question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances
we have been given that From the top of a cliff 92 m high,
the angle of depression of a buoy is 200. we
have to Calculate to the nearest metre, the distance of the buoy from the foot of the cliff.
question no 7 , heights and distances , ICSE board, ML aggarwal
Consider AB as the cliff whose height is 92 m
C is buoy making depression angle of 200
We know that
∠ACB = 200
Take CB = x m
In a right angle triangle ABC
cot θ = BC/AB
Substituting the values
cot 200 = x/92
By cross multiplication
x = 92 × cot 200
So we get
x = 92 × 2.7475
x = 252.7700 m
Hence, the distance of the buoy from the foot of the cliff is 252.77 m.