The angular elevation of a tower from a point is 30 . As we more 100 m nearer to the base of the tower, the angle of elevation become `60^(@)` . Find the height of the tower and the distance of the first point fromt he base of the tower .

Answer

To solve the problem step by step, we will use trigonometric relationships in right triangles formed by the tower and the points of observation.

### Step 1: Define the Variables
Let:
- \( h \) = height of the tower
- \( x \) = distance from the first point to the base of the tower

### Step 2: Set Up the First Triangle
From the first point (let's call it point A), the angle of elevation to the top of the tower is \( 30^\circ \). We can use the tangent function:
\[ \tan(30^\circ) = \frac{h}{x} \]
We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so:
\[ \frac{1}{\sqrt{3}} = \frac{h}{x} \]
This gives us our first equation:
\[ h = \frac{x}{\sqrt{3}} \quad \text{(Equation 1)} \]

### Step 3: Set Up the Second Triangle
After moving 100 meters closer to the tower (to point B), the angle of elevation becomes \( 60^\circ \). The new distance from point B to the base of the tower is \( x - 100 \). Using the tangent function again:
\[ \tan(60^\circ) = \frac{h}{x - 100} \]
We know that \( \tan(60^\circ) = \sqrt{3} \), so:
\[ \sqrt{3} = \frac{h}{x - 100} \]
This gives us our second equation:
\[ h = \sqrt{3}(x - 100) \quad \text{(Equation 2)} \]

### Step 4: Equate the Two Equations for \( h \)
Now we have two expressions for \( h \):
1. \( h = \frac{x}{\sqrt{3}} \)
2. \( h = \sqrt{3}(x - 100) \)

Setting them equal to each other:
\[ \frac{x}{\sqrt{3}} = \sqrt{3}(x - 100) \]

### Step 5: Solve for \( x \)
Cross-multiplying gives:
\[ x = 3(x - 100) \]
Expanding the right side:
\[ x = 3x - 300 \]
Rearranging gives:
\[ 300 = 3x - x \]
\[ 300 = 2x \]
\[ x = 150 \text{ meters} \]

### Step 6: Find the Height of the Tower
Now substitute \( x = 150 \) back into Equation 1 to find \( h \):
\[ h = \frac{150}{\sqrt{3}} = 50\sqrt{3} \text{ meters} \]

### Final Answers
- Height of the tower \( h = 50\sqrt{3} \) meters
- Distance from the first point to the base of the tower \( x = 150 \) meters