The angle of elevation of the top of a vertical tower from two points distance a and b from the base and in the same line with it, are complimentary .If `theta` is the angle subtended at the top of the tower by the line joining these points then sin `theta` =

Options

A. `(a-b)/(sqrt2(a+b))`
B. `(a+b)/(a-b)`
C. `(a-b)/(a+b)`
D. none of these

Answer

To solve the problem, we need to find the value of \( \sin \theta \) where \( \theta \) is the angle subtended at the top of a vertical tower by the line joining two points at distances \( a \) and \( b \) from the base of the tower, with the angles of elevation from these points being complementary.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let the height of the tower be \( h \).
- Let point \( A \) be at distance \( a \) from the base of the tower and point \( B \) be at distance \( b \) from the base.
- The angle of elevation from point \( A \) is \( \phi \) and from point \( B \) is \( 90^\circ - \phi \) (since they are complementary).

2. **Using Trigonometric Ratios**:
- From point \( A \):
\[ \tan \phi = \frac{h}{a} \quad \text{(1)} \]
- From point \( B \):
\[ \tan(90^\circ - \phi) = \cot \phi = \frac{h}{b} \quad \text{(2)} \]

3. **Relating the Two Equations**:
- From (1), we have:
\[ h = a \tan \phi \]
- From (2), we have:
\[ h = b \cot \phi \]
- Setting the two expressions for \( h \) equal gives:
\[ a \tan \phi = b \cot \phi \]

4. **Using the Identity**:
- Recall that \( \cot \phi = \frac{1}{\tan \phi} \), substituting this in gives:
\[ a \tan^2 \phi = b \]
- Rearranging gives:
\[ \tan^2 \phi = \frac{b}{a} \]

5. **Finding \( h \)**:
- Substitute \( \tan \phi \) back into either equation for \( h \):
\[ h = a \tan \phi = a \sqrt{\frac{b}{a}} = \sqrt{ab} \]

6. **Finding \( \theta \)**:
- The angle \( \theta \) subtended at the top of the tower by the line joining points \( A \) and \( B \) can be expressed as:
\[ \theta = 2\phi - 90^\circ \]

7. **Calculating \( \sin \theta \)**:
- Using the sine subtraction formula:
\[ \sin \theta = \sin(2\phi - 90^\circ) = -\cos(2\phi) \]

8. **Expressing \( \cos(2\phi) \)**:
- Using the double angle formula:
\[ \cos(2\phi) = 1 - 2\sin^2 \phi \]

9. **Finding \( \sin^2 \phi \)**:
- From the triangle formed, we have:
\[ \sin \phi = \frac{h}{\sqrt{h^2 + b^2}} = \frac{\sqrt{ab}}{\sqrt{ab + b^2}} = \frac{\sqrt{ab}}{\sqrt{b(a + b)}} \]
- Thus:
\[ \sin^2 \phi = \frac{ab}{b(a + b)} = \frac{a}{a + b} \]

10. **Substituting Back**:
- Now substituting \( \sin^2 \phi \) into \( \cos(2\phi) \):
\[ \cos(2\phi) = 1 - 2\left(\frac{a}{a + b}\right) = \frac{(a + b) - 2a}{a + b} = \frac{b - a}{a + b} \]

11. **Final Expression for \( \sin \theta \)**:
- Therefore:
\[ \sin \theta = -\cos(2\phi) = -\left(\frac{b - a}{a + b}\right) = \frac{a - b}{a + b} \]

### Final Answer:
\[ \sin \theta = \frac{a - b}{a + b} \]