The height of broken part of tree from ground is 5.569m.
Justification:
Let BD is a tree of height 12 m.
<u>Suppose it got bent at a point C and let the part CD take the position CA, meeting the ground at A</u>.
i.e., CD = AC = h m
<u>Broken part makes 60° angle from ground</u>
So, ∠BAC = 60°
<u>Now, height of remaining part of tree</u> = (12 – h)m.
In right angled ∆ABC,
sin 60° = BC/AC
⇒ √3/2 = (12 - h)/h
⇒ √3h = 2(12 – h)
⇒ √3h = 24 – 2h
⇒ √3h + 2h = 24
⇒ h(√3 + 2) = 24
⇒ h(1.732 + 2) = 24
⇒ h(3.732) = 24
⇒ h = 24/3.732 = 6.4308 m
<u>Hence, height of broken tree from ground</u>
⇒ BC = 12 – h
⇒ 12 – 6.4308 = 5.569m
<u>Hence, tree is broken 5.569 m from ground</u>.
<u>Note</u>: See attached picture.
Answer:
a) -6
b) 9
c) 27
Step-by-step explanation:

Answer:
7
Step-by-step explanation:
43.96 divided by 3.14= 14 (diameter)
14 divided by 3= 7 (radius)
You reverse what you did, so first you go 5 units down which makes the coordinates (9,1) then you move 3 units left which then is (6,1). Then you go 4 units up which lets you end out at (6,5).
For this problem, we use the approach of ratio and proportion. Assuming that the given ratio of 444 days per 230 km is constant all throughout, we can determine the number of days or distance as long as one of the two is given. In this case, the solution is as follows:
444 days/230 km = 161616 days/distance
Distance = 83,720 km