A 15 m tall house on the bank of a river is facing a chimney directly opposite to it on the other bank. The angle of depresion from the roof of the house to the base of the chimney is 30 degree and the angle of elevation from the base of the house to the top of the chimney is 60 degree. Find the approximate width of the river and the height of the chimney.
(a)24 m and 47 m
(b)20 m and 25 m
(c)25 m and 41 m
(d)26 m and 45 m

Options

A. 24 m and 47 m
B. 20 m and 25 m
C. 25 m and 41 m
D. 26 m and 45 m

Answer

To solve the problem, we need to find the width of the river and the height of the chimney using the given angles of depression and elevation.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a house that is 15 m tall.
- The angle of depression from the roof of the house to the base of the chimney is 30 degrees.
- The angle of elevation from the base of the house to the top of the chimney is 60 degrees.

2. **Drawing a Diagram**:
- Draw a right triangle where:
- Point A is the top of the house (15 m above the ground).
- Point B is the base of the house (0 m).
- Point C is the base of the chimney on the opposite bank.
- Point D is the top of the chimney.
- The angle of depression from A to C is 30 degrees.
- The angle of elevation from B to D is 60 degrees.

3. **Finding the Width of the River (BC)**:
- In triangle ABC (where A is the top of the house, B is the base of the house, and C is the base of the chimney):
- The height from A to B is 15 m (height of the house).
- The angle of depression to C is 30 degrees.
- Using the tangent function:
\[ \tan(30^\circ) = \frac{BC}{AB} \]
Here, \(AB = 15\) m and \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\).
- Therefore:
\[ \frac{1}{\sqrt{3}} = \frac{BC}{15} \]
\[ BC = 15 \cdot \frac{1}{\sqrt{3}} = \frac{15}{\sqrt{3}} \approx 8.66 \text{ m} \]

4. **Finding the Height of the Chimney (CD)**:
- In triangle BCD (where B is the base of the house, C is the base of the chimney, and D is the top of the chimney):
- The angle of elevation to D is 60 degrees.
- Using the tangent function:
\[ \tan(60^\circ) = \frac{CD}{BC} \]
Here, \(\tan(60^\circ) = \sqrt{3}\).
- Therefore:
\[ \sqrt{3} = \frac{CD}{BC} \]
\[ CD = BC \cdot \sqrt{3} = \left(\frac{15}{\sqrt{3}}\right) \cdot \sqrt{3} = 15 \text{ m} \]
- The total height of the chimney (AD) is:
\[ AD = AB + CD = 15 + 15 = 30 \text{ m} \]

5. **Final Calculations**:
- The width of the river (BC) is approximately 8.66 m.
- The height of the chimney (AD) is approximately 30 m.

### Conclusion:
- The approximate width of the river is 8.66 m, and the height of the chimney is 30 m.