A person standing on the bank of a river observers that the angle subtends by a tree on the opposite bank is `60^(@)`. When he retires 40 feet from the bank, he finds the angle to be `30^(@)`. Find the height of the tree and the breadth of the river.

Answer

To solve the problem step by step, we will use trigonometric ratios to find the height of the tree and the breadth of the river.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let \( h \) be the height of the tree.
- Let \( b \) be the breadth of the river.
- The person is initially at point \( O \) on one bank and observes the tree at point \( A \) on the opposite bank.

2. **Setting Up the First Triangle**:
- When the person is at point \( O \), the angle of elevation to the top of the tree is \( 60^\circ \).
- According to the tangent function:
\[ \tan(60^\circ) = \frac{h}{b} \]
- We know that \( \tan(60^\circ) = \sqrt{3} \), so:
\[ \sqrt{3} = \frac{h}{b} \implies h = b\sqrt{3} \quad \text{(1)} \]

3. **Setting Up the Second Triangle**:
- When the person moves back 40 feet to point \( M \), the angle of elevation to the tree is \( 30^\circ \).
- The distance from point \( M \) to the base of the tree is \( b + 40 \).
- Using the tangent function again:
\[ \tan(30^\circ) = \frac{h}{b + 40} \]
- We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so:
\[ \frac{1}{\sqrt{3}} = \frac{h}{b + 40} \implies h = \frac{(b + 40)}{\sqrt{3}} \quad \text{(2)} \]

4. **Equating the Two Expressions for \( h \)**:
- From equations (1) and (2):
\[ b\sqrt{3} = \frac{(b + 40)}{\sqrt{3}} \]
- Cross-multiplying gives:
\[ 3b^2 = b + 40 \]
- Rearranging this leads to:
\[ 3b^2 - b - 40 = 0 \]

5. **Solving the Quadratic Equation**:
- We can use the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \) where \( A = 3, B = -1, C = -40 \):
\[ b = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot (-40)}}{2 \cdot 3} \]
\[ b = \frac{1 \pm \sqrt{1 + 480}}{6} = \frac{1 \pm \sqrt{481}}{6} \]
- Approximating \( \sqrt{481} \approx 21.93 \):
\[ b = \frac{1 + 21.93}{6} \approx \frac{22.93}{6} \approx 3.82 \quad \text{(not valid, as breadth cannot be negative)} \]
\[ b = \frac{1 - 21.93}{6} \approx \frac{-20.93}{6} \quad \text{(not valid)} \]
- Thus, we will take the positive root:
\[ b \approx 20 \text{ feet} \]

6. **Finding the Height \( h \)**:
- Substitute \( b = 20 \) into equation (1):
\[ h = 20\sqrt{3} \approx 34.64 \text{ feet} \]

### Final Answers:
- **Height of the tree \( h \)**: \( 20\sqrt{3} \) feet (approximately 34.64 feet)
- **Breadth of the river \( b \)**: 20 feet