A pole of height h stands in the centre of a circular platform in the centre of a circular field. Another pole of equal height is at a point on the boundary of the field. The angles of elevation of the top of the first pole from the bottom and top of the second pole are respectively `alpha and beta`. Height of the platform from the ground is

Options

A. `(h cot alpha)/(cot beta-cot alpha)`
B. `(h cot beta)/(cot alpha-cot beta)`
C. `cot alpha-cot beta`
D. None of these

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript and derive the height of the platform from the ground.

### Step 1: Understand the Geometry
We have two poles of equal height \( h \). One pole is at the center of a circular platform, and the other is at the boundary of the circular field. Let’s denote:
- The height of the first pole (center pole) as \( h \).
- The height of the second pole (boundary pole) as \( h \).
- The height of the platform from the ground as \( x \).
- The horizontal distance from the center of the field to the boundary where the second pole is located as \( p \).

### Step 2: Set Up the Angles of Elevation
From the bottom of the second pole (boundary pole), the angle of elevation to the top of the first pole (center pole) is \( \alpha \). From the top of the second pole, the angle of elevation to the top of the first pole is \( \beta \).

### Step 3: Write the Tangent Relationships
1. From the bottom of the second pole (point E) to the top of the first pole (point A):
\[ \tan(\beta) = \frac{h + x}{p} \]
This is because the height from the bottom of the second pole to the top of the first pole is \( h + x \) and the horizontal distance is \( p \).

2. From the top of the second pole (point D) to the top of the first pole (point A):
\[ \tan(\alpha) = \frac{x}{p} \]
This is because the height from the top of the second pole to the top of the first pole is \( x \) and the horizontal distance remains \( p \).

### Step 4: Solve for \( x \)
Now we have two equations:
1. \( \tan(\beta) = \frac{h + x}{p} \) (Equation 1)
2. \( \tan(\alpha) = \frac{x}{p} \) (Equation 2)

From Equation 2, we can express \( p \) in terms of \( x \):
\[ p = \frac{x}{\tan(\alpha)} \]

### Step 5: Substitute \( p \) in Equation 1
Substituting \( p \) from Equation 2 into Equation 1:
\[ \tan(\beta) = \frac{h + x}{\frac{x}{\tan(\alpha)}} \]
This simplifies to:
\[ \tan(\beta) = \frac{(h + x) \tan(\alpha)}{x} \]
Cross-multiplying gives:
\[ x \tan(\beta) = (h + x) \tan(\alpha) \]
Expanding this:
\[ x \tan(\beta) = h \tan(\alpha) + x \tan(\alpha) \]
Rearranging gives:
\[ x \tan(\beta) - x \tan(\alpha) = h \tan(\alpha) \]
Factoring out \( x \):
\[ x (\tan(\beta) - \tan(\alpha)) = h \tan(\alpha) \]
Thus, we find \( x \):
\[ x = \frac{h \tan(\alpha)}{\tan(\beta) - \tan(\alpha)} \]

### Step 6: Express in Terms of Cotangent
Using the identity \( \tan(\theta) = \frac{1}{\cot(\theta)} \):
\[ x = \frac{h \cdot \frac{1}{\cot(\alpha)}}{\frac{1}{\cot(\beta)} - \frac{1}{\cot(\alpha)}} \]
This simplifies to:
\[ x = \frac{h \cot(\beta)}{\cot(\alpha) - \cot(\beta)} \]

### Final Result
The height of the platform from the ground is:
\[ x = \frac{h \cot(\alpha)}{\cot(\beta) - \cot(\alpha)} \]