Two poles of equal heights are standing opposite to each other on either side of the road which is 80m wide, From a point P between them on the road, the angle of elevation of the top of one pole is 60 and the angle of depression from the top of another pole at P is 30 . Find the height of each pole and distance of the point P from the poles.
Solution
Let AB and CD be the equal poles; and BD be the width of the road.
We have,
∠AOB = 60° and ∠COD = 60°
In ΔAOB,
`tan 60° = (AB)/(BO)`
`⇒ sqrt(3) = (AB)/ (BO)`
`⇒ BO = (AB) / sqrt(3)`
Also, in ΔCOD,
`tan 30° = (CD)/(DO)`
`⇒ 1/ sqrt(3) = (CD)/(DO)`
`⇒ DO = sqrt(3)CD`
As , BD = 80
⇒ BO +DO = 80
`⇒ (AB) /sqrt(3) + sqrt(3) CD = 80`
`⇒ (AB) /sqrt(3) + sqrt(3) AB = 80` (Given : AB - CD)
`⇒AB (1/sqrt(3) + sqrt(3) ) = 80`
`⇒ AB ((1+3)/sqrt(3) )=80`
`⇒ AB (4/ sqrt(3) ) =80`
`⇒ AB = (80+sqrt(3))/4`
`⇒ AB = 20 sqrt(3) m`
`Also , BO = (AB)/ sqrt(3) = (20sqrt(3))/sqrt(3) = 20m`
So, DO = 80 - 20 = 60m
Hence, the height of each pole is 20 `sqrt(3)`m and point P is at a distance of 20 m from left pole ad 60 m from right pole.
