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Sagot :

Msm555idn

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

View image Msm555
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