The height of the second building H = h + 24 tan θ = \(\frac {opp}{adj}\) tan 30\(^o = \frac {h}{x}\) \(\implies\frac{\sqrt 3}{3} = \frac {h}{x}\) \(\implies x = \sqrt 3 = 3h\) \(\implies x = \frac {3h}{\sqrt 3}\) ....(i) tan 60\(^o = \frac {24}{x}\) \(\implies\sqrt 3 = \frac {24}{x}\) \(\implies x\sqrt 3 = 24\) \(\implies x = \frac {24}{\sqrt 3}\) ....(ii) Equate equation (i) and (ii) \(\implies \frac {3h}{\sqrt 3} = \frac {24}{\sqrt 3}\) \(\implies\) 3h = 24 \(\implies h = \frac {24}{3}\) = 8m ∴The height of the second building = 8 + 24 = 32m