Let AB be a 5-meter-high house and CD be a palm tree. From the rooftop point A, the base D and the top C of the palm tree are seen at angles of depression and elevation of 30° and 60°, respectively. From point A, a line parallel to BD intersects CD at point E. Since AE || BD, ∠ADB = alternate angle ∠EAD = 30° From right-angled triangle ABD: \[ \tan 30^\circ = \frac{AB}{BD} = \frac{5}{BD} \] \[ \frac{1}{\sqrt{3}} = \frac{5}{BD} \Rightarrow BD = 5\sqrt{3} \] Also, AE = BD = \(5\sqrt{3}\) From right-angled triangle AEC: \[ \tan 60^\circ = \frac{CE}{AE} = \frac{CE}{5\sqrt{3}} \] \[ \sqrt{3} = \frac{CE}{5\sqrt{3}} \Rightarrow CE = 15 \] AEC: \[ \tan 60^\circ = \frac{CE}{AE} = \frac{CE}{5\sqrt{3}} \] \[ \sqrt{3} = \frac{CE}{5\sqrt{3}} \Rightarrow CE = 15 \] Now, \[ CD = CE + ED = CE + AB = 15 + 5 = 20 \] Now, \[ CD = CE + ED = CE + AB = 15 + 5 = 20 \] ∴ The height of the∴ The height of the palm tree CD = palm tree CD = 20 meters And the distance between the house and the20 meters And the distance between the house and the palm tree BD = palm tree BD = \(5\sqrt{3}\) meters\(5\sqrt{3}\) meters.