a 1.3 tall boy is standing at some distance from a 40m tall building the angle of elevation from his eyes to the top of the building increase from 30• to 60• as he moves towards the building find the distance
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in ∆BGC
tan60°=38.7/x...........(40-1.3)
√3=38.7/x
√3x=38.7
x=12.9√3
in ∆ABC
tan30°=38.7/x+Y
1/√3=38.7/x+Y
Y=25.8√3
AC=FD
x+Y=FD
12.9√3+25.8√3=38.7√3
hope it helps you
tan60°=38.7/x...........(40-1.3)
√3=38.7/x
√3x=38.7
x=12.9√3
in ∆ABC
tan30°=38.7/x+Y
1/√3=38.7/x+Y
Y=25.8√3
AC=FD
x+Y=FD
12.9√3+25.8√3=38.7√3
hope it helps you
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