Answer: Height of the second building = (12 + 5√3) m ≈ 20.66 m
Let the first building be AB = 12 m and the second CD. Draw AE ⊥ CD; then AE = BD = 15 m and ED = AB = 12 m. \(\angle CAE=30^\circ\).
\[\tan30^\circ=\frac{CE}{AE}\Rightarrow CE=\frac{15}{\sqrt3}=5\sqrt3\]
\[CD=CE+ED=12+5\sqrt3\approx12+8.66=20.66\ \text{m}\]
Note: the textbook answer key prints 12 + 15√3; with an elevation of 30° over a width of 15 m the rise is 15 tan 30° = 5√3, so the height is 12 + 5√3 m.