Question

The angle of elevation of the top of a to a tower 30m high from the foot of another tower in the same plane is 60°, and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower.

Answer 1

Let the first tower be AB and the second tower be CD.
Given: AB = 30m
​angle ACB = 60o
angle DBC = 30o
Now, in right triangle ABC,
tan 60o = AB/BC
√3 = 30/BC
BC = 30/√3  
=> BC = 30√3/3            (Rationalising)
BC = 10√3m
Now in right triangle BCD,
tan 30o = CD/DB
1/√3 = CD/10√3
CD√3 = 10√3
CD = 10√3/√3
=> CD = 10m
​Therefore, the distance between the two towers is 10√3m (i.e. BC) and the height of the other tower is 10m
Hope this helps.:)

Mark me as a Brainliest.

Answer 2

DISTANCE BETWEEN THE TWO TOWERS=60M
HEIGHT OF SECOND TOWER IS 60/ROOT 3 M