From the top of a 7m high building, the angle of elevation of the the top of a cable tower is 60 and the angleof a depression of its foot is 45 .Determine the height of tower
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Let AB be the tower of lenght 7cm and DE be the cable tower
Also AB=CD where ç lies on DE.
Angle EAC=Angle of elevation=60
Angle CAD=angle of depression=45=angle ADB
Tan 60= EC/AC
√3=x/AC
AC=x/√3
=x√3/3
Also, in triangle ABD-
Tan45=AB*3/x√3
1=3AB/x√3
x√3/3 =AB
x√3/3=7
x=21/√3
x=21*√3/√3*√3
x= 21√3/3
x=7√3
Also AB=CD where ç lies on DE.
Angle EAC=Angle of elevation=60
Angle CAD=angle of depression=45=angle ADB
Tan 60= EC/AC
√3=x/AC
AC=x/√3
=x√3/3
Also, in triangle ABD-
Tan45=AB*3/x√3
1=3AB/x√3
x√3/3 =AB
x√3/3=7
x=21/√3
x=21*√3/√3*√3
x= 21√3/3
x=7√3
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