A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is `60^0dot` When he moves 40 metres away from the bank, he finds the angle of elevation to be `30^@.` Find the height of the tree and the width of the river.

Explanation

Let AB be the tree and AC of the river.
Let C and D be the two positions of the person.
Then, `angleACB = 60^(@), angleADB = 30^(@), angleDAB = 90^(@)` and `CD = 30 m`.
Let AB = h metres and AC = x metres.
From right `DeltaCAB` we have
`(AC)/(AB) = cot 60^(@) = 1/(sqrt(3))`
`rArr x/h =1/(sqrt(3)) rArr x = h/(sqrt(3)) "......."(i)`
`rArr (x+30)/(h) = sqrt(3) rArr x = sqrt(3) h - 30`.
Equating the value of x from (i) and (ii) we get
`h/(sqrt(3)) = sqrt(3) h - 30 rArr h = 3h - 30 sqrt(3)`
`rArr 2h = 30 sqrt(3) rArr h = 15sqrt(3) = 15 xx 1.732 =25.98`.
Putting `h = 15sqrt(3)` in (i), we get `x = (15sqrt(3))/(sqrt(3)) = 15`.
Hence, the height of the treeis `25.98` m and the width of the river is `15` metres.