The two palm trees are of equal heights and are standing opposite each other on either side of the river, which is 80 m wide. From a point O between them on the river the angles of elevation of the top of the trees are 60° and 30°, respectively. Find the height of the trees and the distances of the point O from the trees.

OR

The angles of depression of the top and bottom of a building 50 meters high as observed from the top of a tower are 30° and 60° respectively. Find the height of the tower, and also the horizontal distance between the building and the tower.

Let BD = river

AB = CD = palm trees = h

BO = x

OD = 80 - x

In ΔABO,

tan60°= h/x

√3 = h/x -----------------------(1)

h = √3x

In ΔCDO,

tan 30°= h/(80 - x)

1/√3 = h/(80 - x) ---------------------(2)

Solving (1) and (2), we get

x = 20

h = √3x = 34.6

the height of the trees = h = 34.6m

BO = x = 20m

DO = 80 – x = 80 – 20 = 60m

OR

Let AB = Building of height 50m°

RT = tower of height= h m

BT = AS = x m

AB = ST = 50 m

RS = TR – TS = (h - 50) m

In ΔARS, tan30°= RS/AS

1/√3 = (h-50)/x -------------(1)

In ΔRBT, tan60°=RT/BT

√3 = h/x --------------(2)

Solving (1) and (2), we get

h= 75

from (2)

x = h/√3

= 75/√3

= 25√3

Hence, height of the tower = h = 75m

Distance between the building and the tower = x = 25√3 = 43.25m