A tree is broken by the wind. The top of that tree struck the ground a

1.	A tree is broken by the wind. The top of that tree struck the ground at an angleof 30Â°& at a distance of 30 m from the root. Find the height of the whole tree.
Answer» 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. Hit like if you find it useful!

A tree is broken by the wind. The top of that tree struck the ground at an angleof 30Â°& at a distance of 30 m from the root. Find the height of the whole tree.

Answer»

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.

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