The upper part of a tree is broken by the wind so that the broken top makes an angle of 60° with the ground and touches the ground at a point 25 metres from the foot of the tree. What was the original height of the tree (in metres) before it broke?

Introduction / Context:
This is another broken-tree trigonometry question. The upper part of the tree bends over and touches the ground, forming a right triangle with the ground and the remaining upright part of the trunk. We use the given angle and distance along the ground to compute the original height of the tree.

Given Data / Assumptions:

• The tree was initially vertical.
• The top breaks and touches the ground at a point 25 m from the base.
• The broken top makes an angle of 60° with the ground.
• The broken part remains straight and attached at the breaking point.
• We must find the total original height of the tree.

Concept / Approach:
The broken upper part is the hypotenuse of a right triangle whose base is 25 m and angle at the ground is 60°. From this, we compute the length of the broken part (hypotenuse) using cos(60°). The vertical projection of this hypotenuse is the height at which the tree broke. Adding this break height to the hypotenuse length returns the original height of the tree, because before breaking that top piece stood vertically above the break point.

Step-by-Step Solution:
Let L be the length of the broken top of the tree.Base distance from tree foot to top's contact point = 25 m.Angle between broken part and ground = 60°.cos(60°) = base / hypotenuse = 25 / L.cos(60°) = 1 / 2 ⇒ 1 / 2 = 25 / L ⇒ L = 25 * 2 = 50 m.Vertical height from ground to point of break = L * sin(60°).sin(60°) = √3 / 2 ⇒ break height = 50 * (√3 / 2) = 25√3 m.Original height H = break height + length of broken part = 25√3 + 50.Approximate: √3 ≈ 1.732 ⇒ 25√3 ≈ 25 * 1.732 = 43.3.So H ≈ 43.3 + 50 = 93.3 m.

Verification / Alternative check:
We can verify by reconstructing the triangle: if the hypotenuse is 50 m and angle 60°, the base should be 50 * cos(60°) = 25 m, matching the given base distance. The vertical leg is 50 * sin(60°) = 25√3, so adding this to the hypotenuse gives the original height. Both the geometry and the arithmetic are consistent.

Why Other Options Are Wrong:
84.14 m, 98.25 m, 120.24 m, 75 m: These values do not correspond to 25√3 + 50 and do not satisfy the right-triangle relations with base 25 m and angle 60°. Plugging any of them back fails the trigonometric relationships.

Common Pitfalls:
Common mistakes include treating 25 m as the length of the broken top instead of the horizontal base, or forgetting to add the stump height and the broken part. Others misapply sin(60°) and cos(60°), or assume that the original height equals only one of the legs. A quick diagram and correct identification of the hypotenuse make the process much clearer.

Final Answer:
The original height of the tree was approximately 93.3 m.