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There is a small island in between a river 100 metres wide. A tall tree stands on the island. P and Q are points directly opposite to each other on the two banks and in the line with the tree. If the angles of elevation of the top of the tree from P and Q are 300 and 450 respectively, find the height of the tree.
There is a small island in between a river 100 metres wide. A tall tree stands on the island. P and Q are points directly opposite to each other on the two banks and in the line with the tree. If the angles of elevation of the top of the tree from P and Q are 300 and 450 respectively, find the height of the tree.
ML Agarwal Heights and distances class 10 Maths
There is a small island in between a river 100 metres wide. A tall tree stands on the island. P and Q are points directly opposite to each other on the two banks and in the line with the tree. If the angles of elevation of the top of the tree from P and Q are 300 and 450 respectively, find the height of the tree.

Solution:

Width of the river PQ = 100 m

B is the island and AB is the tree on it

Angles of elevation from A to P and Q are 300 and 450

Consider AB = h

PB = x

BQ = 100 – x

In right triangle APB

tan θ = AB/PB

Substituting the values

tan 300 = h/x

So we get

1/√3 = h/x

x = √3h ….. (1)

In right triangle ABQ

tan θ = AB/BQ

Substituting the values

tan 450 = h/ (100 – x)

So we get

1 = h/ (100 – x)

h = 100 – x ….. (2)

Using both the equations

h = 100 – √3h

By further calculation

h + √3h = 100

So we get

(1 + 1.732) h = 100

h = 100/ 2.732

Multiply and divide by 1000

h = (100 × 1000)/ 2732

h = 100000/ 2732

h = 36.6

Hence, the height of the tree is 36.6 m.

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