The angle of elevation of the top of a tower 30 m 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 and also the height of the tower.

Explanation

Let the distance between the two towers = AB = x m and height of the other tower = PA = h m
Given that height of the tower = QB = 30 M and `angleQAB=60^(@), anglePBA=30^(@)`
Now, in `DeltaQAB, tan60^(@)= (QB)/(AB)=30/x`
`rArr sqrt(3)=30/x`
`therefore x=(30)/(sqrt(3)). sqrt(3)/sqrt(3) = (30sqrt(3))/(sqrt(3))= (30sqrt(3))/(3) = 10sqrt(3)`m
and in `DeltaPBA,`
`tan30^(@)= (PA)/(AB) = h/x`
`rArr 1/sqrt(3)= h/(10sqrt(3))`
`rArr h=10m`
Hence, the required distance and height are `10sqrt(3)`m and 10 m respectively.