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.

Answer :


https://gs-post-images.grdp.co/user_files/63094/15047/images/extra-9_files/dew.gif

Given:

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

Solution:


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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