a man 1.5meter is standing at a distance of 10√3 meter from a building.if the angle of elevation from his eyes to the top of the building is 60°find the height of the building
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FINAL ANSWER IS 31.5 metres.
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Let AB=CD= EQ = 1.5 m
Let dist. covered by boy is BD=x
Let BD=AC=x
Let DQ=CE=y
In fig. PE=30-1.5=28.5m
Angles are 30° and 60°
In ∆ PAE
tan 30 = PE/AE
1/√3=28.5/x+y
28.5√3=x+y –(1)
In ∆PCE
tan60=PE/CE
√3=28.5/y
y=28.5/√3
Put in (1)
x+28.5/√3=28.5√4
x=(28.5×3-28.5)/√3
= (85.5-28.5)/√3
=57/√3
Rationalize
=57/√3×√3/√3
=57√3/3
=19√3
Let dist. covered by boy is BD=x
Let BD=AC=x
Let DQ=CE=y
In fig. PE=30-1.5=28.5m
Angles are 30° and 60°
In ∆ PAE
tan 30 = PE/AE
1/√3=28.5/x+y
28.5√3=x+y –(1)
In ∆PCE
tan60=PE/CE
√3=28.5/y
y=28.5/√3
Put in (1)
x+28.5/√3=28.5√4
x=(28.5×3-28.5)/√3
= (85.5-28.5)/√3
=57/√3
Rationalize
=57/√3×√3/√3
=57√3/3
=19√3
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