A boat is sailing towards a lighthouse of height 20√3 m at a certain speed. The angle of elevation of the top of the lighthouse changes from 30° to 60° in 10 seconds. What is the time taken (in seconds) by the boat to reach the lighthouse from its initial position?

Given : CD = $$20\sqrt3$$ m and time taken to reach B from A = 10 seconds

To find : Time taken to reach D from A

Solution : In $$\triangle$$ BCD,

=> $$tan(60^\circ)=\frac{CD}{BD}$$

=> $$\sqrt3=\frac{20\sqrt3}{BD}$$

=> $$BD=20$$ m

Similarly, in $$\triangle$$ ACD,

=> $$tan(30^\circ)=\frac{CD}{AD}$$

=> $$\frac{1}{\sqrt3}=\frac{20\sqrt3}{20+AB}$$

=> $$AB+20=60$$

=> $$AB=60-20=40$$ m

=> Speed of boat (while travelling from A to B) = distance/time

= $$\frac{40}{10}=4$$ m/s

=> AD = BD + AB = 20 + 40 = 60 m

$$\therefore$$ Time taken to reach D from A = $$\frac{60}{4}=15$$ seconds

=> Ans - (B)