Question

A man on on the top of a lighthouse observes a boat coming towards it if takes 10 minutes for the angle of depression to change is from 30to 45 degrees then the time taken by the boat to reach the lighthouses is _______

Answer 1

Answer:

assume that let AB be the lighthouse of height h and DC be the distance of boat where its angle of depression changed from 30 to 45 , whereas let BD be the total distance travelled by boat at base of lighthouse

it is given that time taken by boat  is 10 mins

therefore time taken in seconds will be 600 sec (1 min = 60 sec)

let speed of boat be x metre/seconds

so distance travelled from d to c = 600*x = 600x  (distance=speed*time)

let BC be y , now in Δ ABD , tan 30 = AB/BD

⇒ 1/√3 = h/600x +y ---------------(i)

Now in  Δ ABC, tan 45 =AB/y

⇒ 1= h/y ---------------(ii)

dividing (ii) by (i)

so h/y / h/600x+y = 1 /  1/√3

⇒h*(600x+y)/ h*y = √3/1

therefore 600x+y / y = √3

so 600 x + y = √3 y

600x= √3y-y

600x= (√3-1)y

600x/(√3-1) = y or y=600x/(√3-1)*(√3+1)/ (√3+1)      (rationalise it)

600x(√3+1) / 2

=300x(√3+1)=y

                    or

          300x√3 + 300x = y---------------------(iii)

now total distance = BD

⇒ 600x +y = 600x + 300x + 300x√3

⇒900x + 300x√3

⇒300x(3+√3)

so 300x(3+√3)= BD or total distance travelled by boat at the base of lighthouse

so time taken by boat to reach the lighthouse = distance/ speed

= 300x(3+√3) / x   (speed of boat = xmetre/sec)

⇒300(3+√3)

300(4.732)      (√3=1.732)

so time taken is 300 (4.732) sec

time taken in minutes= 300(4.732)/60  (1 sec = 1/60 min)

=23.66 minutes = ANSWER