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 concepts related to heights and distances.

### Step 1: Understand the scenario and draw a diagram
- Let the height of the tree be \( h \).
- Let the width of the river be \( b \).
- The person is initially at point A (on the bank of the river), observing the tree at point C across the river.
- The angle of elevation to the top of the tree from point A is \( 60^\circ \).
- The person then moves back 40 feet to point D. The angle of elevation to the top of the tree from point D is \( 30^\circ \).

### Step 2: Set up the first triangle (triangle ABC)
Using the tangent function in triangle ABC:
\[ \tan(60^\circ) = \frac{h}{b} \]
From trigonometric values, we know that \( \tan(60^\circ) = \sqrt{3} \). Therefore:
\[ \sqrt{3} = \frac{h}{b} \implies h = b \sqrt{3} \quad \text{(Equation 1)} \]

### Step 3: Set up the second triangle (triangle DBC)
In triangle DBC, the person is now at point D, which is 40 feet away from the bank:
\[ \tan(30^\circ) = \frac{h}{b + 40} \]
We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \). Therefore:
\[ \frac{1}{\sqrt{3}} = \frac{h}{b + 40} \implies h = \frac{(b + 40)}{\sqrt{3}} \quad \text{(Equation 2)} \]

### Step 4: Equate the two expressions for \( h \)
From Equation 1 and Equation 2, we have:
\[ b \sqrt{3} = \frac{(b + 40)}{\sqrt{3}} \]
Multiplying both sides by \( \sqrt{3} \):
\[ 3b = b + 40 \]
Rearranging gives:
\[ 3b - b = 40 \implies 2b = 40 \implies b = 20 \text{ feet} \]

### Step 5: Find the height of the tree \( h \)
Now substituting \( b = 20 \) back into Equation 1:
\[ h = b \sqrt{3} = 20 \sqrt{3} \text{ feet} \]

### Final Results
- The height of the tree \( h \) is \( 20 \sqrt{3} \) feet.
- The width of the river \( b \) is \( 20 \) feet.