If the angles of elevation of the top of a tower from two places situated at distances 21 m and x m from the base of the tower are `45^(@) and 60^(@)` respectively, then what is the value of x?

Options

A. `7 sqrt3 m `
B. ` 7- sqrt3 m `
C. ` 7 + sqrt 3m `
D. ` 14 `

Answer

To solve the problem, we will use the concept of angles of elevation and the properties of right triangles.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a tower (let's denote its height as AB) and two points (C and D) from which the angles of elevation to the top of the tower are given. The distances from the base of the tower (point B) to points C and D are 21 m and x m, respectively. The angles of elevation from these points are 45° and 60°.

2. **Setting Up the First Triangle (ABC)**:
From point C (21 m away from the tower), the angle of elevation is 45°.
- We know that \( \tan(45°) = 1 \).
- In triangle ABC, we can write:
\[ \tan(45°) = \frac{AB}{BC} = \frac{h}{21} \]
- Since \( \tan(45°) = 1 \), we have:
\[ 1 = \frac{h}{21} \implies h = 21 \text{ m} \]

3. **Setting Up the Second Triangle (ABD)**:
From point D (x m away from the tower), the angle of elevation is 60°.
- We know that \( \tan(60°) = \sqrt{3} \).
- In triangle ABD, we can write:
\[ \tan(60°) = \frac{AB}{BD} = \frac{h}{x} \]
- Substituting the height \( h = 21 \) m:
\[ \sqrt{3} = \frac{21}{x} \]

4. **Solving for x**:
- Rearranging the equation gives:
\[ x = \frac{21}{\sqrt{3}} \]
- To rationalize the denominator:
\[ x = \frac{21 \sqrt{3}}{3} = 7 \sqrt{3} \text{ m} \]

5. **Final Answer**:
The value of \( x \) is \( 7 \sqrt{3} \) m.