The upper part of a tree, broken over by the wind. makes an angle of `45^(@)` with the ground , and the distance from the root to the point where the top of the tree touches the ground. is 1.5 m. What was the height of the tree before it was broken ?

Answer

To find the height of the tree before it was broken, we can follow these steps:

### Step 1: Understand the Problem
We have a tree that has been broken by the wind. The broken part makes an angle of \(45^\circ\) with the ground, and the distance from the root to the point where the top of the tree touches the ground is \(1.5\) meters.

### Step 2: Set Up the Diagram
Let:
- Point A be the root of the tree.
- Point B be the point where the tree is broken.
- Point C be the point where the top of the tree touches the ground.

We know:
- The angle \( \angle BCA = 45^\circ \)
- The distance \( AC = 1.5 \) m (the distance from the root to the point where the top touches the ground).

### Step 3: Use Trigonometric Ratios
In triangle \( ABC \):
- We can use the tangent function since we know the angle and the opposite side.

Using the tangent function:
\[ \tan(45^\circ) = \frac{AB}{AC} \]
Since \( \tan(45^\circ) = 1 \):
\[ 1 = \frac{AB}{1.5} \]
Thus, we find:
\[ AB = 1.5 \text{ m} \]

### Step 4: Find the Length of BC
Now, we need to find the length of \( BC \). We can use the sine function:
\[ \sin(45^\circ) = \frac{BC}{AB} \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ \frac{1}{\sqrt{2}} = \frac{BC}{1.5} \]
Rearranging gives us:
\[ BC = 1.5 \cdot \frac{1}{\sqrt{2}} = \frac{1.5}{\sqrt{2}} = \frac{1.5 \sqrt{2}}{2} = 0.75\sqrt{2} \text{ m} \]

### Step 5: Calculate the Total Height of the Tree
The total height of the tree before it was broken is the sum of \( AB \) and \( BC \):
\[ \text{Height of the tree} = AB + BC = 1.5 + 0.75\sqrt{2} \]

Calculating \( 0.75\sqrt{2} \):
\[ 0.75 \cdot 1.414 \approx 1.06 \text{ m} \]
Thus,
\[ \text{Height of the tree} \approx 1.5 + 1.06 = 2.56 \text{ m} \]

### Final Answer
The height of the tree before it was broken is approximately \( 2.56 \) meters.

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