The upper part of a vertical tree is broken by the wind and the broken top still attached to the trunk touches the ground, making an angle of 60 degrees with the ground. The horizontal distance between the root of the tree and the point where the top touches the ground is 25 metres. Assuming the tree was originally straight, what was the height of the tree in metres?

Introduction / Context:
This is a standard application of trigonometry to a practical situation. A tree has broken such that its top still remains attached and touches the ground some distance from the base. The broken part makes a known angle with the ground. We must determine the original height of the tree by modelling the situation as a right angled triangle and using the given angle and distance along the ground.

Given Data / Assumptions:

• The tree originally stood vertically on level ground.
• The top portion was broken but stayed attached and now touches the ground.
• The broken top makes an angle of 60 degrees with the horizontal ground.
• The horizontal distance between the base (root) of the tree and the point where the top touches the ground is 25 metres.
• The tree broke at some point above the ground and not at the base.

Concept / Approach:
Let the original height of the tree be H metres and the height at which it broke be h metres. Then the length of the broken top is H − h. When the top touches the ground, the broken part and the ground form a right angled triangle where the broken part is a slant side and the base of the tree is another vertex. The distance between the root and the point of contact on the ground is the horizontal base of a right triangle with included angle 60 degrees between the broken top and the ground. Using trigonometry, we can express the height of the break in terms of the distance and angle, and then add the length of the top to obtain the original height.

Step-by-Step Solution:
Step 1: Consider triangle formed by the root O, the break point B, and the ground contact point C. The right angle is at O, and the angle at C between the broken top BC and the ground is 60°. Step 2: Let OC, the horizontal distance, be 25 metres. In triangle OBC, angle at C is 60° and angle at O is 90°. Therefore angle at B is 30°. Step 3: Using trigonometry at point C, the vertical height OB is opposite to angle 60° in triangle OBC, and BC is the broken part. Step 4: Using tan 60° = OB / OC, we can write √3 = OB / 25, so OB = 25√3 metres. Step 5: The broken part BC is the hypotenuse. Use sin 60° = OB / BC. Thus √3 / 2 = (25√3) / BC, giving BC = 50 metres. Step 6: The original height H is OB + BC = 25√3 + 50. Using √3 ≈ 1.732, H ≈ 25 × 1.732 + 50 ≈ 43.3 + 50 = 93.3 metres.

Verification / Alternative check:
We can verify the geometry by computing OC from the triangle. Using cos 60° = OC / BC gives 1 / 2 = 25 / BC if OC is 25 metres. Then BC = 50 metres, consistent with the earlier calculation. With OB = 25√3 and BC = 50, Pythagoras also holds, because BC² = OB² + OC² becomes 50² = (25√3)² + 25², that is 2500 = 1875 + 625, confirming the right angled triangle.

Why Other Options Are Wrong:
Options 84.14, 98.25, and 120.24 represent different supposed heights that do not satisfy the correct trigonometric relationships when we reconstruct the triangle. Option 80 is a simple approximate value but fails when checked with tan 60° and sin 60° for the specific 25 metre base. Only 93.3 metres is consistent with all conditions of the problem.

Common Pitfalls:
Some learners mistakenly consider the 25 metres as the length of the broken part instead of the horizontal distance. Others forget that the original height is the sum of the unbroken upright part and the broken top. Confusing sine and cosine for the angle at the ground is another common issue. Careful drawing of the diagram and labelling of sides helps in avoiding these mistakes.

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