Question

solve this question

Answer 1

AB= height of the boy- 1.5m

CD= height of building -30m

CG= 30-1.5=28.5m

in triangle CGF

CG/FG= tan 60degree

28.5/FG= √3

FG= 28.5/√3×√3/√3=28.5√3/3

in triangle CGA

CG/GA= tan30degree

28.5/GA= 1/√3

GA= 28.5√3

Now,

BE= AG= FG

28.5√3-28.5√2/3

28.5×3√3-28.5√3/3

85.5√3-28.5√3/3

57√3/3

19√3

therefore, the distance he walked towards the building bis19√3

Answer 2

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