Length of the vertical pole = 6m (Given)
Shadow of the tree = 4 m (Given)
Let Height of tower = h m
Length of shadow of the tower = 28 m (Given)
In ΔABC and ΔDEF,
∠C = ∠E (angular elevation of sum)
∠B = ∠F = 90°
∴ ΔABC ~ ΔDEF (By AA similarity criterion)
∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)
∴ 6/h = 4/28
⇒ h = 6×28/4
⇒ h = 6 × 7
⇒ h = 42 m
Hence, the height of the tower is 42 m.
