Question

A 2m tall boy is standing out at a distance from 29 m tall builiding. Angle of elevation from his eyes of the top of the building increases from 30 to 60 as he walks towards the building. Find the distance he walked towards the building. Ans is 18root3

Answer 1

AF = 29m
BE = 2m
AC = AF - BE
AC = 29m - 2m
AC = 27m
In triangle DAC,
We know that, tan 60degree = AC/DC
We also know that, tan 60degree = root 3
Root 3 = AC/DC
Root 3 = 27/DC
DC = 1/root 3*27
=27/root 3 --- Equation 1
Tn triangle BAC
We know that, tan 30degree = AC/BC
We know that, tan 30degree = 1/root 3
AC/BC = 1/root 3
27/BC = 1/root 3
27 root 3 = BC --- Equation 2
Distance he walked = BC - DC --- Equation 3
Substitute equation 1 and equation 2 in equation 3
Distance he walked = 27 root 3 - 27/root 3
27 root 3 - 27/3*root3
27 root 3 - 9 root 3
Distance he walked = 18 root 3

Answer 2

I hope this is helpful