Q11 of 107 Page 12

The shadow of a tower, when the angle of elevation of the sun is 45°, is found to be 10 m longer than when it was 60°. Find the height of the tower.

Let the height of the tower = h (m)



Let the point of 60° elevation is (m) away from the foot of the tower.


In ∆ABC,


tan 45° =


1 =


1 =


h = 10+ ----(1)


In ∆ABD,


tan 60° =


√3 =


h = √3


= ---------(2)


From eqn. (2) in eqn. (1)


h =


h - = 10


= 10


= 10√3


⇒


h= ⇒ h=


⇒ 15+5√3 ⇒ 23.66 m.


Therefore height of the tower is 23.66 m.


More from this chapter

All 107 →
9

A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff of height 5 metres. At a point on the plane, the angles of elevation of the bottom and the top of the flag-staff are respectively 30° and 60°. Find the height of the tower.

 10

A person observed the angle of elevation of the top of a tower as 30°. He walked 50 m towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as 60°. Find the height of the tower.

 12

A parachutist is descending vertically and makes angles of elevation of 45° and 60° at two observing points 100 m apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of the point where he falls on the ground from the just observation point.

 13

On the same side of a tower, two objects are located. When observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 150 m, find the distance between the objects.