#### Question

A person standing on the bank of river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60°. When he moves 40 m away from the bank, he finds the angle of elevation to be 30°. Find the height of the tree and width of the river. `(sqrt 3=1.73)`

#### Solution

Let AB = height of the tower = h metres

In the right angled ΔABC and right angled ΔABD,

`tan60^@=h/(BC)=sqrt(3) rArr BC=h/sqrt(3)`

`tan30^@=h/(BD)=1/sqrt(3) rArr BD=hsqrt(3)`

`Now, BD – BC = 40`

`hsqrt(3)-h/sqrt(3)=40`

`(3h-h)/sqrt3=40`

`2h=40sqrt3`

`h=20sqrt3 m`

In right angled ΔABC,

`tan30°=1/sqrt3`

`(BC)/h=1/sqrt3`

`(BC)/(20sqrt3)=1/sqrt3`

`BC=b=20m`

Width of the river = BC = 20 m

Thus, the height of the tree is `20 sqrt3` metres and width of the river is 20 metres.

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#### APPEARS IN

Solution A person standing on the bank of river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60° Concept: Heights and Distances.