A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is `60^(@)`, when he retreats 40m 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, we will use trigonometric principles. Let's denote the height of the tree as \( h \) and the breadth of the river as \( d \).

### Step 1: Set up the scenario
1. Let the point where the person is standing on the bank of the river be point \( A \).
2. The tree on the opposite bank is point \( B \).
3. The person retreats 40 meters to point \( C \).
4. The angle \( \angle ABC = 60^\circ \) when he is at point \( A \).
5. The angle \( \angle CBA = 30^\circ \) when he is at point \( C \).

### Step 2: Use the tangent function
From point \( A \):
\[ \tan(60^\circ) = \frac{h}{d} \]
From point \( C \):
\[ \tan(30^\circ) = \frac{h}{d + 40} \]

### Step 3: Calculate the tangent values
We know:
\[ \tan(60^\circ) = \sqrt{3} \quad \text{and} \quad \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

### Step 4: Set up the equations
From the first equation:
\[ \sqrt{3} = \frac{h}{d} \implies h = d\sqrt{3} \quad \text{(1)} \]
From the second equation:
\[ \frac{1}{\sqrt{3}} = \frac{h}{d + 40} \implies h = \frac{d + 40}{\sqrt{3}} \quad \text{(2)} \]

### Step 5: Equate the two expressions for \( h \)
From equations (1) and (2):
\[ d\sqrt{3} = \frac{d + 40}{\sqrt{3}} \]

### Step 6: Cross-multiply to eliminate the fraction
\[ d\sqrt{3} \cdot \sqrt{3} = d + 40 \]
\[ 3d = d + 40 \]

### Step 7: Solve for \( d \)
\[ 3d - d = 40 \implies 2d = 40 \implies d = 20 \, \text{meters} \]

### Step 8: Substitute \( d \) back to find \( h \)
Using \( d = 20 \) in equation (1):
\[ h = 20\sqrt{3} \approx 20 \times 1.732 \approx 34.64 \, \text{meters} \]

### Final Answers
- Height of the tree \( h \approx 34.64 \, \text{meters} \)
- Breadth of the river \( d = 20 \, \text{meters} \)