Question

using trigonometry find AB and BC​

Answer 1

Answer:

I)

in ∆abd

tan30°=ab/bd

1/√3= ab/bc+20(as bd =bc+cd)

bc+20=√3.ab (EQ 1)

now in ∆abc

tan40°=ab/bc

1=ab/bc

ab=bc (eq2)

now putting the of bc value in EQ I

ab+20=√3.ab

20=√3.ab-ab

20= ab(√3-1)

20/√3-1=ab

20(√3+1)/√3-1(√3+1)=ab

20√3+1/3-1=ab

20√3+1/2=ab

10√3+1=ab

now as ab = bc (from eq 2)

bc=10√3+1

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