An object is observed from the points A, B and C lying in a horizontal straight line which passes directly underneath the object. The angular elevation at B is twice that at A and at C three times that at A. If AB=a, BC=b, then the height of the object is

Options

A. `(3a)/(2b) sqrt((a+b) (3b-a))`
B. `(3b)/(2a) sqrt((a+b) (3a-b))`
C. `(a)/(2b) sqrt((a+b) (3b-a))`
D. None of the above

Answer

To solve the problem, we need to find the height of the object (let's denote it as \( h \)) based on the given conditions regarding the angles of elevation from points A, B, and C.

### Step-by-Step Solution:

1. **Define the Angles of Elevation**:
- Let the angle of elevation at point A be \( \theta \).
- Then, the angle of elevation at point B is \( 2\theta \) (twice that of A).
- The angle of elevation at point C is \( 3\theta \) (three times that of A).

2. **Set Up the Relationships Using Trigonometry**:
- From point A:
\[ \tan(\theta) = \frac{h}{d_A} \]
where \( d_A \) is the horizontal distance from A to the point directly below the object.
- From point B:
\[ \tan(2\theta) = \frac{h}{d_B} \]
where \( d_B \) is the horizontal distance from B to the point directly below the object.
- From point C:
\[ \tan(3\theta) = \frac{h}{d_C} \]
where \( d_C \) is the horizontal distance from C to the point directly below the object.

3. **Express Distances in Terms of \( a \) and \( b \)**:
- Given that \( AB = a \) and \( BC = b \), we can express:
\[ d_B = d_A - a \]
\[ d_C = d_B + b = d_A - a + b \]

4. **Substituting Distances**:
- Substitute \( d_B \) and \( d_C \) into the equations:
\[ \tan(2\theta) = \frac{h}{d_A - a} \]
\[ \tan(3\theta) = \frac{h}{d_A - a + b} \]

5. **Using the Tangent Double and Triple Angle Formulas**:
- Recall the formulas:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
\[ \tan(3\theta) = \frac{3\tan(\theta) - \tan^3(\theta)}{1 - 3\tan^2(\theta)} \]

6. **Set Up the Equations**:
- From the equations for \( \tan(2\theta) \) and \( \tan(3\theta) \):
\[ \frac{2h}{d_A - a} = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
\[ \frac{3h - h\tan^3(\theta)}{1 - 3\tan^2(\theta)} = \frac{h}{d_A - a + b} \]

7. **Solve for Height \( h \)**:
- Rearranging these equations will give you expressions for \( h \) in terms of \( d_A \), \( a \), and \( b \).
- After simplification, you will find that:
\[ h = \frac{2a \tan(\theta)}{1 - \tan^2(\theta)} \]
and similarly for the other equations.

8. **Final Expression**:
- After solving the equations, you will arrive at a final expression for \( h \) in terms of \( a \) and \( b \).