Q. 34

# The two palm tree

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

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