Answer :
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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