Question

A tree is broken by the wind.the top struck the ground at angle of 300 and at a distance of 30 metres from the root of the tree.Find the whole height of the tree.(use √3=1.732)

Answer 1

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.