Q18 of 25 Page 9

As observed from the top of a lighthouse, 200 m above the sea level, the angle of depression of a ship sailing towards it changes from 30° to 45°. Determine the distance travelled by the ship during the period of observation.

To find:


The distance travelled by ship.


Formula Used:



Explanation:



Let A and B be two positions of the ship.


Let d be the distance travelled by the ship i.e. AB = d metres


PO = 200 metres


∴ ∠OAP = 30° and ∠OBP = 45°


In Δ OPB,




⇒ BP = 200 m


In ΔOPA,




⇒ d + BP = 200√3


⇒ d + 200 = 200√3


⇒ d = 200√3 – 200


⇒ d = 200(√3 – 1)


⇒ d = 200(1.732 – 1)


d = 146.4 m


Hence distance travelled by ship is 146.4 m.


More from this chapter

All 25 →
16

A window of house is h metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite site of the lane are found to be α and β respectively. Prove that the height of the other house is h (1+ tanα cotβ).

 17

The pillars of equal height stand on either side of a roadway which is 150 m wide. From a point on the roadway between the pillars, the elevation of the top of the pillar are 60° and 30° . Find the height of the pillars and the position of the point.

 19

If the angle of elevation of a cloud from a point 200 m above a lake is 30° and the angle of depression of its reflection in the lake is 60°, then find the height of the cloud above the lake.

 20

Two poles are ‘a’ metres apart and the height of one is double of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then find the height of the smaller pole.