Question

answer this plzx ...............,.........

Answer 1

let CD be 1.5 tall boy AB be the height of the building,let cf be the distance he walked towards the building
AE=AB-BE
30-1.5=28.5
in right triangle AEF
$tan 60 = ae \div ef$
√3 =28.5÷ef
If =28.5÷√3
in right triangle AEC
tan 30 =ae÷ec
1/√3= 28.5/ef + FC
28.5√3=28.5÷√3+fc
fc =28.5√3 -28.5÷√3
28.5√3×√3-28.5÷√3
= 19√3