Q. 75.0( 2 Votes )
There is a small island in the middle of a 100 m wide river and atall tree stands on the island. P and Q are points directly opposite eachother on the two banks, and in line with the tree. If the anglesof elevation of the top of the tree from P and Q are respectively 30º and45º, find the height of the tree.
Width of the river, w = PQ = 100 m
Angle of elevation of the top of tree from P, θ = 30°
Angle of elevation of the top of tree from Q, θ = 45°
To find: Height of the tree
Let PQ be the width of the river and RS be the height of the tree on the island.
In right angled Δ PRS,
x = RS cot 30°
x = RS × √ 3
x = √ 3 × RS ……………….(i)
In right angled Δ RSQ,
SQ = RS cot 45°
(100 – x) = RS
x = 100 - RS ………………. (ii)
Equating (i) and (ii) we have:
√ 3 × RS = 100 – RS
2.73 RS = 100
RS = 36.63 m
Therefore the height of the tree is 36.63 m.
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