Let AB be the lighthouse, and from point A at the top of the lighthouse, the angles of depression to the masts of two ships located along the same straight line are 60° and 30°, at points C and D respectively. So, ∠EAC = 60° and ∠EAD = 30° (assuming AE || BD) Given: BC = distance to the nearer ship = 150 meters ∠ADB = ∠EAD = 30° and ∠ACB = ∠EAC = 60° From right-angled triangle △ABC: \[ \tan 60^\circ = \frac{AB}{BC} = \frac{AB}{150} \Rightarrow \sqrt{3} = \frac{AB}{150} \Rightarrow AB = 150\sqrt{3} \] From right-angled triangle △ABD: \[ \tan 30^\circ = \frac{AB}{BD} = \frac{150\sqrt{3}}{BD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{150\sqrt{3}}{BD} \Rightarrow BD = 150\sqrt{3} \times \sqrt{3} = 450 \] ✅ Therefore, the farther ship is **450 meters** away from the lighthouse, and the height of the lighthouse is **150√3 meters**.