The shadow of a tower standing on a level plane is found to be 50 m longer when the Sun's elevation is `30^(@)` than when it is `60^(@).` The height of the tower is

Options

A. A. 25 m
B. B. `25 sqrt 3 m `
C. C. ` 50 m `
D. D. None of these

Answer

To solve the problem step by step, we will use the properties of right triangles and trigonometric ratios.

### Step 1: Define the variables
Let:
- \( H \) = height of the tower
- \( X \) = length of the shadow when the sun's elevation is \( 60^\circ \)
- The length of the shadow when the sun's elevation is \( 30^\circ \) will be \( X + 50 \) meters.

### Step 2: Set up the equations using trigonometric ratios
1. For the angle of elevation \( 60^\circ \):
\[ \tan(60^\circ) = \frac{H}{X} \]
We know that \( \tan(60^\circ) = \sqrt{3} \), so:
\[ \sqrt{3} = \frac{H}{X} \implies H = \sqrt{3} \cdot X \quad \text{(Equation 1)} \]

2. For the angle of elevation \( 30^\circ \):
\[ \tan(30^\circ) = \frac{H}{X + 50} \]
We know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so:
\[ \frac{1}{\sqrt{3}} = \frac{H}{X + 50} \implies H = \frac{X + 50}{\sqrt{3}} \quad \text{(Equation 2)} \]

### Step 3: Set the two equations for \( H \) equal to each other
From Equation 1 and Equation 2, we can set them equal to each other:
\[ \sqrt{3} \cdot X = \frac{X + 50}{\sqrt{3}} \]

### Step 4: Clear the fraction by multiplying both sides by \( \sqrt{3} \)
\[ 3X = X + 50 \]

### Step 5: Solve for \( X \)
Rearranging the equation gives:
\[ 3X - X = 50 \implies 2X = 50 \implies X = 25 \text{ meters} \]

### Step 6: Substitute \( X \) back to find \( H \)
Now, substitute \( X \) back into Equation 1 to find \( H \):
\[ H = \sqrt{3} \cdot 25 = 25\sqrt{3} \text{ meters} \]

### Final Answer
The height of the tower is \( 25\sqrt{3} \) meters.

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