From a point 12 m above the water level, the angle of elevation of the top of a hill is `60^(@)` and the angle of depression of the base of the hill is `30^(@)`. What is the height (in m)of the hill?

Options

A. `36`
B. `48sqrt3`
C. `48`
D. `36sqrt3`

Answer

To find the height of the hill, we can break the problem down into two right triangles formed by the angles of elevation and depression from the point above the water level.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Height of the point above water level (point A) = 12 m
- Angle of elevation to the top of the hill (point B) = 60°
- Angle of depression to the base of the hill (point C) = 30°

2. **Draw the Diagram:**
- Draw a vertical line representing the height of 12 m (point A).
- From point A, draw a line at a 60° angle upwards to point B (top of the hill).
- From point A, draw a line at a 30° angle downwards to point C (base of the hill).
- Let the horizontal distance from point A to point C be represented as AC.

3. **Calculate AC (Horizontal Distance to Base of the Hill):**
- In triangle AOC (where O is the point directly below A on the water level):
- Using the tangent of the angle of depression:
\[ \tan(30°) = \frac{12}{AC} \]
- We know that \(\tan(30°) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{12}{AC} \]
- Cross-multiplying gives:
\[ AC = 12\sqrt{3} \text{ m} \]

4. **Calculate the Height of the Hill (BC):**
- In triangle AOB:
- Using the tangent of the angle of elevation:
\[ \tan(60°) = \frac{h}{AC} \]
- We know that \(\tan(60°) = \sqrt{3}\):
\[ \sqrt{3} = \frac{h}{12\sqrt{3}} \]
- Cross-multiplying gives:
\[ h = 12\sqrt{3} \cdot \sqrt{3} = 12 \cdot 3 = 36 \text{ m} \]

5. **Calculate the Total Height of the Hill (BD):**
- The total height of the hill (BD) is the sum of the height of point A (12 m) and the height of the hill (BC):
\[ BD = AC + BC = 12 + 36 = 48 \text{ m} \]

### Final Answer:
The height of the hill is **48 m**.