A tree is broken by the wind. The top struck the ground at an angle of 30º and at a distance of 30 m from the root. Find the whole height of the tree. ( √3 = 1.73)

5. A tree is broken by the wind. The top struck the ground at an angle of 30º and at a distance of 30 m from the root. Find the whole height of the tree. ( √3 = 1.73)

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.