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.
