Answer:
Solution given:
CD=18m
In ∆ ACD
Tan 35=opposite/adjacent
Tan 35=CD/AC
AC=18/Tan35
AC=25.7m
again
In ∆ BCD
Tan 60=opposite/adjacent
Tan 60=CD/BC
BC=18/tan60
BC=10.4
we have
AC=AB+BC
AB=25.7-10.4
AB=15.3
Distance between A and B point is 15.3 metre
