# 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. 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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