A tower standing on a horizontal plane subtends a certain angle at a point 160 m apart from the foot of the tower. On advancing 100 m towards if, the tower is found to subtend an angle twice as before. The height of the tower is -

Options

A. 80 m
B. 100 m
C. 160 m
D. 200 m

Answer

To solve the problem, we will follow these steps:

1. **Understand the Problem**: We have a tower of height \( H \). A point \( C \) is located 160 m from the foot of the tower, and at this point, the angle subtended by the tower is \( \alpha \). When we move 100 m closer to the tower (to point \( D \)), the angle becomes \( 2\alpha \).

2. **Set Up the Diagram**:
- Let \( A \) be the top of the tower, \( B \) be the foot of the tower, \( C \) be the initial point (160 m from \( B \)), and \( D \) be the point after moving 100 m closer (60 m from \( B \)).
- The height of the tower is \( H \).

3. **Using Tangent for Angles**:
- From point \( C \):
\[ \tan(\alpha) = \frac{H}{160} \]
- From point \( D \):
\[ \tan(2\alpha) = \frac{H}{60} \]

4. **Using the Double Angle Formula**:
- The double angle formula for tangent is:
\[ \tan(2\alpha) = \frac{2\tan(\alpha)}{1 - \tan^2(\alpha)} \]
- Substituting \( \tan(\alpha) \):
\[ \tan(2\alpha) = \frac{2\left(\frac{H}{160}\right)}{1 - \left(\frac{H}{160}\right)^2} \]

5. **Setting the Two Expressions for \(\tan(2\alpha)\) Equal**:
- From point \( D \):
\[ \frac{H}{60} = \frac{2\left(\frac{H}{160}\right)}{1 - \left(\frac{H}{160}\right)^2} \]

6. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ H(1 - \left(\frac{H}{160}\right)^2) = \frac{2H \cdot 60}{160} \]
- Simplifying:
\[ H - \frac{H^3}{25600} = \frac{120H}{160} \]
- Further simplifying:
\[ H - \frac{H^3}{25600} = \frac{3H}{4} \]

7. **Rearranging the Equation**:
- Rearranging gives:
\[ H - \frac{3H}{4} = \frac{H^3}{25600} \]
- This simplifies to:
\[ \frac{H}{4} = \frac{H^3}{25600} \]

8. **Multiplying Both Sides by 25600**:
- Multiply both sides by 25600:
\[ 6400H = H^3 \]

9. **Rearranging to Form a Polynomial**:
- Rearranging gives:
\[ H^3 - 6400H = 0 \]
- Factoring out \( H \):
\[ H(H^2 - 6400) = 0 \]

10. **Finding the Roots**:
- This gives us \( H = 0 \) or \( H^2 = 6400 \).
- Thus, \( H = 80 \) (since height cannot be negative).

So, the height of the tower is \( H = 80 \) m.