An object is observed from three points A,B,C in the same horizontal plane passing through the base of the object. The angle of elevation at B is twice and at C is thrice that at A. if AB=a, BC=b, then the height of the object is `(a)/(2b)sqrt((a+b)(3b-a))`.

Answer

To solve the problem step-by-step, we will use trigonometric relationships and the properties of triangles. Here’s the detailed solution:

### Step 1: Define the Angles
Let the angle of elevation at point A be \( \theta \).
Then, the angle of elevation at point B is \( 2\theta \) and at point C is \( 3\theta \).

### Step 2: Establish Relationships for Height
From point A, the height \( H \) of the object can be expressed using the tangent function:
\[ H = d_A \tan(\theta) \]
where \( d_A \) is the horizontal distance from point A to the base of the object.

From point B, the height can be expressed as:
\[ H = d_B \tan(2\theta) \]
From point C, it can be expressed as:
\[ H = d_C \tan(3\theta) \]

### Step 3: Express Distances
Let:
- \( AB = a \)
- \( BC = b \)

Assuming the distances from the object to points A, B, and C are as follows:
- \( d_A = d \)
- \( d_B = d - a \)
- \( d_C = d - a - b \)

### Step 4: Use Trigonometric Identities
Using the double angle and triple angle formulas for tangent:
\[ \tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)} \]
\[ \tan(3\theta) = \frac{3\tan(\theta) - \tan^3(\theta)}{1 - 3\tan^2(\theta)} \]

### Step 5: Set Up the Equations
From the equations of height:
1. \( H = d \tan(\theta) \)
2. \( H = (d - a) \tan(2\theta) \)
3. \( H = (d - a - b) \tan(3\theta) \)

### Step 6: Equate Heights
Equating the heights from points A and B:
\[ d \tan(\theta) = (d - a) \tan(2\theta) \]
Equating the heights from points A and C:
\[ d \tan(\theta) = (d - a - b) \tan(3\theta) \]

### Step 7: Solve for \( H \)
Using the relationships from the tangent equations, we can express \( H \) in terms of \( a \) and \( b \):
1. Rearranging gives us:
\[ H = \frac{d \tan(\theta)}{1 - \tan^2(\theta)} \]
2. Substituting the values of \( \tan(2\theta) \) and \( \tan(3\theta) \) into the equations and simplifying will lead us to the final expression for \( H \).

### Final Expression
After simplifying the expressions, we arrive at:
\[ H = \frac{a}{2b} \sqrt{(a + b)(3b - a)} \]

### Conclusion
Thus, the height of the object is given by:
\[ H = \frac{a}{2b} \sqrt{(a + b)(3b - a)} \]