From the top and bottom of a 40 meter high tower, the angle of elevation of top of a light house are respectively `30^(@) and 60^(@)`. Find the height of the light house. Also find the distance between top of light house and foot of tower.

Answer

To solve the problem step by step, we will use the concepts of trigonometry, specifically the tangent function for right-angled triangles.

### Step 1: Understand the problem
We have a tower of height 40 meters. From the top of the tower, the angle of elevation to the top of the lighthouse is 30 degrees, and from the bottom of the tower, the angle of elevation is 60 degrees. We need to find the height of the lighthouse and the distance between the top of the lighthouse and the foot of the tower.

### Step 2: Define the variables
- Let \( AB \) be the height of the lighthouse.
- Let \( AE \) be the height from the base of the tower to the base of the lighthouse.
- Let \( BE \) be the height of the tower, which is 40 meters.
- Thus, \( AB = AE + BE \) or \( AB = AE + 40 \).

### Step 3: Set up the triangles
1. From the bottom of the tower (point C) to the top of the lighthouse (point A), we have triangle \( ABC \) where:
- Angle \( ACB = 60^\circ \)
- \( AB \) is the opposite side (height of the lighthouse).
- \( BC \) is the adjacent side (distance from the tower to the lighthouse).

2. From the top of the tower (point B) to the top of the lighthouse (point A), we have triangle \( ABE \) where:
- Angle \( ABE = 30^\circ \)
- \( AE \) is the opposite side (height from the base of the tower to the base of the lighthouse).
- \( BE \) is the adjacent side (which is the same as \( BC \)).

### Step 4: Use the tangent function
For triangle \( ABC \):
\[ \tan(60^\circ) = \frac{AB}{BC} \]
\[ \sqrt{3} = \frac{AB}{BC} \quad \text{(since } \tan(60^\circ) = \sqrt{3}\text{)} \]
Thus, we can express \( BC \) as:
\[ BC = \frac{AB}{\sqrt{3}} \]

For triangle \( ABE \):
\[ \tan(30^\circ) = \frac{AE}{BC} \]
\[ \frac{1}{\sqrt{3}} = \frac{AE}{BC} \quad \text{(since } \tan(30^\circ) = \frac{1}{\sqrt{3}}\text{)} \]
Thus, we can express \( BC \) as:
\[ BC = \sqrt{3} \cdot AE \]

### Step 5: Set the equations equal
Since both expressions represent \( BC \):
\[ \frac{AB}{\sqrt{3}} = \sqrt{3} \cdot AE \]

### Step 6: Substitute \( AB \)
Substituting \( AB = AE + 40 \):
\[ \frac{AE + 40}{\sqrt{3}} = \sqrt{3} \cdot AE \]

### Step 7: Clear the fraction
Multiply through by \( \sqrt{3} \):
\[ AE + 40 = 3AE \]

### Step 8: Solve for \( AE \)
Rearranging gives:
\[ 40 = 3AE - AE \]
\[ 40 = 2AE \]
\[ AE = 20 \text{ meters} \]

### Step 9: Calculate \( AB \)
Now substituting back to find \( AB \):
\[ AB = AE + 40 = 20 + 40 = 60 \text{ meters} \]

### Step 10: Find the distance \( BC \)
Using \( BC = \frac{AB}{\sqrt{3}} \):
\[ BC = \frac{60}{\sqrt{3}} = 20\sqrt{3} \text{ meters} \]

### Step 11: Conclusion
- The height of the lighthouse is **60 meters**.
- The distance between the top of the lighthouse and the foot of the tower is **20√3 meters**.