Answer:
79.9 ft
Solution,
Let AB is the Building and MA is Boy's height.
Suppose BC = X feet
In ∆MBC
[tex]tan \: 53 = \frac{ma + ab}{bc} \\ tan \: 53 = \frac{106}{x} \\ tan \: 53 \times x = 106 \\ x = \frac{106}{tan \: 53} \\ x = 79.9 \: ft[/tex]
Hope this helps...
Good luck on your assignment...