Q12 of 35 Page 219

As observed from a point 60 m above a lake, the angle of elevation of an advertising balloon has measure 30 and from the same point the angle of depression of the image of the balloon in the lake has measure 60. Calculate the height of the balloon above the lake.

Let BE be the surface of the lake.



A is a point 60 m above a lake.


F is the image of balloon C in the lake.


Horizontal line AD intersects CE in D.


∠CAD = 30⁰, ∠FAD = 60⁰, AB = 60 m


Let CE = h, BE = l


Then CD = h – 60, DF = h + 60


In ∆ADC,




L = √3(h – 60)……….(1)


In ∆ADF,




√3l = h + 60…………(2)


From equation (1) and (2),


√3[√3(h – 60)] = h + 60


3(h – 60) = h + 60


3h – 180 = h + 60


2h = 240


H = 120 m


The height of balloon above the lake is 120 m.


More from this chapter

All 35 →
10

From a point at the height 100 m above the sea level, the angles of depression of a ship in the sea is found to have measure 30. After some time the angle of depression of the ship has measure 45. Find the distance travelled by the ship during that time interval.

 11

From the top of a 300 m high light – house, the angles of depression of the top and foot of a tower have measure 30 and 60. Find the height of the tower.

 13

Watching from a window 40 m high of a multi – storeyed building, the angle of elevation of the top of a tower is found to have measure 45. The angle of elevation of the top of the same tower from the bottom of the building is found to have measure 60. Find the height of the tower.

 14

Two pillars of equal height stand on either side of a road, which is 100 m wide. The angles of elevation of the top of the pillars have measure 60 and 30 at a point on the road between the pillars. Find the position of the point from the nearest end of a pillars and the height of pillars.