Durga was standing on a railway overbridge that is 5√3 meters high. She observed the engine of a moving passenger train at a depression angle of 30° on one side of the bridge. Two seconds later, she saw the same engine at a depression angle of 60° on the other side of the bridge. Durga's position was vertically above the railway track, which is assumed to be a straight line. Find the speed of the train.

Q.Durga was standing on a railway overbridge that is 5√3 meters high. She observed the engine of a moving passenger train at a depression angle of 30° on one side of the bridge. Two seconds later, she saw the same engine at a depression angle of 60° on the other side of the bridge. Durga's position was vertically above the railway track, which is assumed to be a straight line. Find the speed of the train.

Let’s assume Durga is standing at point A on the BC railway overbridge. She first sees the engine of the train at point S, and after 2 seconds, she sees it at point E. The angles of depression are ∠BAS = 30° and ∠CAE = 60°, respectively. AP = height of the railway overbridge = 5√3 meters Since BC || SE, ∠ASP = alternate angle ∠BAS = 30° and ∠AEP = alternate angle ∠CAE = 60° From right-angled triangle ASP: tan30° = \(\cfrac{AP}{SP} = \cfrac{5√3}{SP}\) ⇒ \(\cfrac{1}{√3} = \cfrac{5√3}{SP}\) ⇒ \(SP = 15\) From right-angled triangle APE: tan60° = \(\cfrac{AP}{PE} = \cfrac{5√3}{PE}\) ⇒ \(\sqrt3 = \cfrac{5√3}{PE}\) ⇒ \(PE = \cfrac{5√3}{√3} = 5\) ∴ Distance covered by the train in 2 seconds = SE = SP + PE = 15 + 5 = 20 meters ∴ Speed of the train = \(\cfrac{20}{2}\) meters/second = \(10\) meters/second