Question

7. The shadow of a tower standing on a level plane is found to be 50 m
longer when sun's elevation is 30° than when it is 60°. Find the height
the tower.​

Answer 1

Given BD-BC = CD = 50m

∴ In ΔABC,

tan 60° = AB/BC

=> AB/√3 = BC ...(1)

∴ In ΔABD,

tan 30° = AB/BD

=> 1/√3 = AB/(BC+CD)

=> AB√3 = AB/√3+50 (From 1)

=> AB√3-AB/√3 = 50

=> 3AB-AB/√3 = 50

=> AB = 50√3/2 = 25√3 = 25*1.732 = 43.30m