The angle of elevation of the tip of a tower from a point on the ground is `45^@`.Moving 21 m directly towards the base of the tower,the angle of elevation changes to `60^@`. What is the height of the tower, to the nearest metre ?

Options

A. 48 m
B. 49 m
C. 50 m
D. 51m

Answer

To solve the problem, we will use trigonometric ratios, specifically the tangent function, which relates the angle of elevation to the height of the tower and the distance from the point of observation to the base of the tower.

### Step-by-Step Solution:

1. **Understand the Problem**:
- Let the height of the tower be \( h \).
- Let the distance from the first point (where the angle of elevation is \( 45^\circ \)) to the base of the tower be \( d \).

2. **Set Up the First Equation**:
- From the first point, the angle of elevation is \( 45^\circ \).
- Using the tangent function:
\[ \tan(45^\circ) = \frac{h}{d} \]
- Since \( \tan(45^\circ) = 1 \), we have:
\[ 1 = \frac{h}{d} \implies h = d \]

3. **Move Towards the Tower**:
- The problem states that the observer moves 21 m towards the base of the tower.
- Therefore, the new distance from the base of the tower is \( d - 21 \).

4. **Set Up the Second Equation**:
- Now, the angle of elevation is \( 60^\circ \).
- Using the tangent function again:
\[ \tan(60^\circ) = \frac{h}{d - 21} \]
- Since \( \tan(60^\circ) = \sqrt{3} \), we have:
\[ \sqrt{3} = \frac{h}{d - 21} \implies h = \sqrt{3}(d - 21) \]

5. **Substitute the First Equation into the Second**:
- From \( h = d \), substitute \( h \) in the second equation:
\[ d = \sqrt{3}(d - 21) \]

6. **Solve for \( d \)**:
- Distributing \( \sqrt{3} \):
\[ d = \sqrt{3}d - 21\sqrt{3} \]
- Rearranging gives:
\[ d - \sqrt{3}d = -21\sqrt{3} \]
\[ d(1 - \sqrt{3}) = -21\sqrt{3} \]
- Thus,
\[ d = \frac{-21\sqrt{3}}{1 - \sqrt{3}} \]

7. **Calculate \( d \)**:
- To simplify, multiply numerator and denominator by the conjugate of the denominator:
\[ d = \frac{-21\sqrt{3}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{-21\sqrt{3}(1 + \sqrt{3})}{1 - 3} = \frac{-21\sqrt{3}(1 + \sqrt{3})}{-2} \]
\[ d = \frac{21\sqrt{3}(1 + \sqrt{3})}{2} \]

8. **Find \( h \)**:
- Now substitute \( d \) back into \( h = d \):
\[ h = \frac{21\sqrt{3}(1 + \sqrt{3})}{2} \]

9. **Calculate \( h \) Numerically**:
- Using \( \sqrt{3} \approx 1.732 \):
\[ h \approx \frac{21 \times 1.732 \times (1 + 1.732)}{2} \approx \frac{21 \times 1.732 \times 2.732}{2} \approx \frac{21 \times 4.732}{2} \approx \frac{99.372}{2} \approx 49.686 \]

10. **Round to the Nearest Metre**:
- Therefore, the height of the tower \( h \) is approximately \( 50 \) m.

### Final Answer:
The height of the tower is **50 meters**.