Answer: Height of the building = 16√3 m (about 27.71 m)
Let AB be the building (height h) and C the boy, with BC = 48 m and \(\angle ACB=30^\circ\). In right ΔABC:
\[\tan30^\circ=\frac{AB}{BC}\Rightarrow\frac1{\sqrt3}=\frac h{48}\]
\[h=\frac{48}{\sqrt3}=\frac{48\sqrt3}{3}=16\sqrt3\ \text{m}\]
With \(\sqrt3\approx1.732\), \(h\approx27.71\) m.