Question

the upper part of the tree broken over by wind makes an angle of 30° with the ground and distance of the root from the point where the top touches the ground is 25m what was the height of tree

Answer 1






Sol. seg AB represents the height of the tree


The tree breaks at point D


∴ seg AD is the broken part of tree which then takes the position of DC

∴  AD = DC

m∠ DCB = 30º

BC = 30 m

In right angled ∆DBC,

tan 30º = side opposite to angle 30º/adjacent side of 30º

∴ tan 30º = BD/BC

∴ 1/√3 = BD/30

∴ BD = 30/√3

∴ BD = (30/√3)×(√3/√3)

∴ BD = 10√3 m

cos 300 = adjacent side of angle 300/Hypotenuse


∴ cos 300 = BC/DC

∴ cos 300 = 30/DC

∴ √3/2  = 30/DC

∴ DC = (30×2)/√3

∴ DC = (60/√3)×(√3/√3)

∴ DC = 20√3 m

AD = DC = 20 √3 m

AB = AD + DB [∵ A - D - B]

∴  AB = 20 3 + 10 3

∴  AB = 30 3 m

∴  AB = 30 × 1.73

∴  AB = 51.9 m


∴  The height of tree is 51.9 m