The upper part of a tree is broken by the strong wind. The top of the tree makes an angle of 308 with the horizontal ground. The distance between the base of the tree and the point where it touches the ground is 10 m. Find the total height of the tree.
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Answer:
Thus, total height of the tree is 10√3 m
Step-by-step explanation:
Hi,
Let the upper part of a tree after broken be AC touching the
ground at A and resting on the horizontal making an angle of
30° with the horizontal.
Consider triangle ABC, where
AC is the broken part.
BC is the remaining part of the tree.
Given ∠BAC = 30°
Given AB = 10 m
tan ∠BAC = BC/AB
⇒ 1/√3 = BC/10
⇒BC = 10/√3.
cos∠BAC = AB/AC
⇒cos 30° = 10/AC
⇒√3/2 = 10/AC
⇒AC = 20/√3
Total Height of the tree = AC + BC
= 20/√3 + 10/√3
= 30/√3
= 10√3.
Thus, total height of the tree is 10√3 m
Hope, it helps !
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Let length of tree before windstorm is BD.
After windstorm the upper part of tree C falls from point C to point A on the ground
. Now, let CD = AC = h2 m AB = 10 m
Broken part makes an angle 60° from the ground.
So, ∠CAB = 60°
From right angled ∆ABC tan 60° = BC/AB ⇒ √3 = h1/10 ⇒ h1 = 10√3 and cos 60° = AB/AC ⇒ 1/2 = 10/h2 ⇒ h2 = 10 × 2 = 20 m
Hence, total length of tree BD = BC + CD = h1 + h1
= 10√3 + 20
= 10 × 1.732 + 20
= 17.32 + 20
= 37.32 m
Hence, height of the tree 37.32 m.
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