A tree is broken by the wind. The top struck the ground at an angle of 45 and the distance 35m from the foot .then find the whole height of the tree before broken. (trignometry)
Answers
Answer:
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Step-by-step explanation:
height of the whole tree. = 30√3 = 51.9 m
Let say initial Height of Tree = T m
Let say it Broken at x m height
Then length of broken part = T - x m
Broken part make an angle of 30° with ground
and touch 30 m away from root
Tan 30 = x / 30
=> 1/√3 = x/30
=> x = 10√3
(T - x)² = x² + 30²
=> ( T - 10√3)² = (10√3)² + 900
=> ( T - 10√3)² = 300 + 900
=> ( T - 10√3)² = 1200
=> ( T - 10√3) = 20√3
=> T = 30√3 = 51.9 m
height of the whole tree. = 30√3 = 51.9 m
the trunk of a tree cracks at a point 10 m above the ground the tree
A straight tree is broken due to thunder storm. The broken part is ...
Step-by-step explanation:
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.
hope it helps!!