During the super cyclone in Orissa when wind blew a coconut tree touched the top of a pole 9 ft. high. The distance of the pole from the tree 12 ft. After some time the tree brokedown and touched the bottom of another pole which is `5sqrt(3)` ft. from the tree. Find the angle from by the broken part of tree with the ground.

Options

A. `30^(@)`
B. `20^(@)`
C. `60^(@)`
D. `45^(@)`

Answer

To solve the problem step by step, we will first analyze the situation and then apply the necessary mathematical concepts.

### Step 1: Understand the Problem
We have a coconut tree that initially touches the top of a pole that is 9 ft high, and the distance from the tree to the pole is 12 ft. When the tree breaks, it touches the ground at a distance of \(5\sqrt{3}\) ft from the tree.

### Step 2: Calculate the Length of the Tree
When the tree was bent, it formed a right triangle with the pole and the ground. We can use the Pythagorean theorem to find the length of the tree (let's denote it as \(L\)).

\[ L^2 = \text{height of the pole}^2 + \text{distance from the tree to the pole}^2 \]
\[ L^2 = 9^2 + 12^2 = 81 + 144 = 225 \]
\[ L = \sqrt{225} = 15 \text{ ft} \]

### Step 3: Set Up the Equation for the Broken Tree
Let \(x\) be the length of the part of the tree that remains standing and \(y\) be the length of the part that fell. We know:
\[ x + y = 15 \quad \text{(1)} \]
The distance from the tree to the point where it touches the ground is given as \(5\sqrt{3}\) ft. Using the Pythagorean theorem again, we have:
\[ x^2 - y^2 = (5\sqrt{3})^2 \]
\[ x^2 - y^2 = 75 \quad \text{(2)} \]

### Step 4: Solve the System of Equations
From equation (1), we can express \(x\) in terms of \(y\):
\[ x = 15 - y \]
Substituting this into equation (2):
\[ (15 - y)^2 - y^2 = 75 \]
Expanding the left side:
\[ 225 - 30y + y^2 - y^2 = 75 \]
This simplifies to:
\[ 225 - 30y = 75 \]
\[ 30y = 225 - 75 \]
\[ 30y = 150 \]
\[ y = 5 \text{ ft} \]
Now substituting back to find \(x\):
\[ x = 15 - y = 15 - 5 = 10 \text{ ft} \]

### Step 5: Find the Angle with the Ground
Now we have a right triangle where the height of the broken part of the tree is \(y = 5\) ft, and the distance from the tree to where it touches the ground is \(5\sqrt{3}\) ft. We need to find the angle \(\theta\) that the broken part of the tree makes with the ground.

Using the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{y}{5\sqrt{3}} = \frac{5}{5\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Step 6: Calculate the Angle
The angle \(\theta\) for which \(\tan(\theta) = \frac{1}{\sqrt{3}}\) is:
\[ \theta = 30^\circ \]

### Final Answer
The angle made by the broken part of the tree with the ground is \(30^\circ\).

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