Math Problem Statement

A tree was broken due to a heavy storm. The broken part of the tree touches the ground, and the distance between the base of the tree and the point where the broken part touches the ground is 8 meters. The broken part makes a 60° angle with the ground. Find the total height of the tree.

Solution

The problem in the image describes a tree that has broken due to a storm. Here is the summary of the problem:

  • The broken part of the tree touches the ground.
  • The distance between the base of the tree (its root) and the point where the broken part touches the ground is 8 meters.
  • The broken part makes an angle of 60° with the ground.

We are asked to find the total height of the tree.

Solution:

This problem can be solved using trigonometry. The scenario describes a right triangle formed by:

  • The vertical part of the tree that remains upright (let's call it h1h_1).
  • The broken part of the tree (let's call its length h2h_2).
  • The distance on the ground between the tree base and the point where the tree touches the ground is given as 8 m.

The broken part forms a hypotenuse in this right triangle, where:

  • The adjacent side (along the ground) is 8 m.
  • The angle formed with the ground is 6060^\circ.

Using the cosine function: cos(60)=Adjacent (ground distance)Hypotenuse (broken tree part)\cos(60^\circ) = \frac{\text{Adjacent (ground distance)}}{\text{Hypotenuse (broken tree part)}} cos(60)=8h2\cos(60^\circ) = \frac{8}{h_2}

Since cos(60)=0.5\cos(60^\circ) = 0.5, we can solve for h2h_2: 0.5=8h20.5 = \frac{8}{h_2} h2=16 metersh_2 = 16 \text{ meters}

Now that we know the length of the broken part of the tree (h2=16h_2 = 16 m), we can find the height of the upright part of the tree using the sine function: sin(60)=Opposite (upright height)Hypotenuse (broken part)\sin(60^\circ) = \frac{\text{Opposite (upright height)}}{\text{Hypotenuse (broken part)}} sin(60)=h116\sin(60^\circ) = \frac{h_1}{16} Since sin(60)=32\sin(60^\circ) = \frac{\sqrt{3}}{2}, we get: 32=h116\frac{\sqrt{3}}{2} = \frac{h_1}{16} h1=16×32=8313.86 metersh_1 = 16 \times \frac{\sqrt{3}}{2} = 8\sqrt{3} \approx 13.86 \text{ meters}

Finally, the total height of the tree is the sum of h1h_1 (the upright part) and h2h_2 (the broken part): Total height=h1+h2=13.86+16=29.86 meters\text{Total height} = h_1 + h_2 = 13.86 + 16 = 29.86 \text{ meters}

Conclusion:

The total height of the tree is approximately 29.86 meters.


Would you like more details or have any questions?

Here are 5 related questions for further exploration:

  1. How would the total height change if the broken part made a 45° angle with the ground?
  2. How does the use of trigonometry help in solving real-world problems like this?
  3. What would be the total height if the ground distance was 10 meters?
  4. How can the Pythagorean theorem be applied in problems involving right triangles?
  5. What is the relationship between the angle and the length of the broken part of the tree?

Tip: Always sketch a diagram when solving geometry or trigonometry problems to help visualize the relationships between angles and sides.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometry
Right Triangles

Formulas

cos(θ) = adjacent / hypotenuse
sin(θ) = opposite / hypotenuse

Theorems

Trigonometric Ratios

Suitable Grade Level

Grades 9-12