A man of a cliff observes a boat at an angle of depression of `30^0` which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be `60^@.` Find the time taken by the boat to reach the shore.

Explanation

Let AB be the cliff with the man at B. the boat approaching the shore at A.
Then,
`angleACB = 30^(@)`and `angleADB = 60^(@)`.
Let `AB = h` metres, `CD =x` metres and `DA = y` metres.
From right `DeltaDAB`, we have
`(AD)/(AB) = cot 60^(@) = 1/(sqrt(3)) rArr y/h = 1/(sqrt(3)) rArr y = h/(sqrt(3)) "........."(i)`
From right `DeltaCAB`, we have
`(AC)/(aB) = cot30^(@) = sqrt(3) rArr (x+y)/(h) = sqrt(3) rArr x+y = hsqrt(3)"........."(ii)`
On substracting (i) from (ii), we get
`x = (hsqrt(3)-h/(sqrt(3))) rArr x = (2h)/(sqrt(3))."............"(iii)`
From (iii) and (i) we get
`x : y = (2h)/(sqrt(3)) : h/(sqrt(3)) = 2:1`,
Let the time taken to cover y units be t minutes. Then,
ratio of distances = ratio of times taken to cover them.
So, `2:1 = 6:t rArr 2/1 = 6/t rArr 2t = 6rArrt= 3`.
Thus, the boat takes 3 mintues to cover the distance DA.
Hence, the total time taken by the boat to roach the shore is `(6+3)` minutes `= 9` minutes.