As observed from the top of a lighthouse, 100 metres high above sea level, the angle of depression of a ship moving directly towards it, changes from 30o to 60o. Determine the distance travelled by the ship during the period of observation.
Let A and B be the two positions of the ship. Let d be the distance travelled by the ship during the period of observation, i.e., AB=d metres.
Suppose that the observer is at the point P. It is given that PC = 100m.
Let h be the distance (in metres) from B to C.
In right triangle PCA,
= cot30o =
d+h = 100 ... (i)
In triangle PCB,
= cot60o =
h = metres
Putting the value of h in (i), we get,
Thus, the distance travelled by the ship from A to B is 115.47 metres (approx.).
You have rated this answer /10
- CBSE Sample Papers for Class 10 Mathematics
- R S Aggarwal and V Aggarwal Textbook Solutions for Class 10 Mathematics
- RD Sharma Textbook Solutions for Class 10 Mathematics
- NCERT Textbook Solutions for Class 10 Mathematics
- CBSE Syllabus for Class 10 Mathematics
- CBSE Previous year papers with Solutions Class 10 Mathematics