The shadow of a tower, when the angle of elevation of the sun is 45° is found to be a 10 m long then when it is 60°. Find the height of the tower.
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Answer:
let AB be the tower with height h m
In ΔABC
tan 45°=
1=
h= 10+x -------(1)
In ΔABD
tan 60° =
√3=
h= x√3 --------(2)
from 1 and 2
10+x=x√3
x√3-x =10
x(√3-1) = 10
x(0.732) = 10
x=
x= 13.66m
height of tower is 13.66m
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