At a point on level ground, the angle of elevation of a vertical tower

is found to be such that its tangent is . On walking 192 metres

towards the tower, the tangent of the angle of elevation is . Find

the height of the tower.


The horizontal distance between two towers is 140 m. The angle of elevation of the top of the first tower when seen from the top of the second tower is 30°. The height of the second tower is 60 m find the height of the first tower.

Asked by Topperlearning User | 30th Nov, 2013, 02:13: AM

Expert Answer:


Let AB be the tower of height h metres. Let AD=x metres, CD=192 metres.

tanα = , tanβ =


tanα = = ……………………. (i)


tanβ = = or x= ……………………… (ii)

Using (ii) in (i)


5= 12h

2880 +20h =36h

16h = 2880 or h= 180

Hence, the height of the tower is 180 metres.


Let AB and CD be two towers of height h m and 60 m respectively.

AC=140m and ÐBDE =300.


tan 30° =


BE == 80.83m

Thus, the height of the first tower is

AB= AE+BE = CD+BE =60+80.83 = 140.83m

Answered by  | 30th Nov, 2013, 04:13: AM