ABC is a triangular park with AB+AC=100 m. a clock tower is situated at the mid-point of BC. The angles of elevation of the top of the tower at A and B are `cot^(-1)" "3*2 and "cosec"^(-1)" "2*6` respectively. The height of the tower is

Options

A. 16 m
B. 25 m
C. 50 m
D. None

Answer

To find the height of the clock tower situated at the midpoint of BC in the triangular park ABC, we will follow these steps:

### Step 1: Understand the Given Information
We know:
- \( AB + AC = 100 \, \text{m} \)
- The angles of elevation from points A and B to the top of the tower are given as:
- \( \alpha = \cot^{-1}(3.2) \)
- \( \beta = \csc^{-1}(2.6) \)

### Step 2: Determine the Height of the Tower
Let the height of the tower be \( h \). The midpoint of BC is denoted as O.

From point A:
- The angle of elevation \( \alpha \) gives us:
\[ OA = h \cdot \cot(\alpha) = h \cdot 3.2 \]

From point B:
- The angle of elevation \( \beta \) gives us:
\[ OB = h \cdot \cot(\beta) \]
To find \( \cot(\beta) \), we first find \( \sin(\beta) \):
\[ \sin(\beta) = \frac{1}{\csc(\beta)} = \frac{1}{2.6} \]
Then, using the Pythagorean identity:
\[ \cos(\beta) = \sqrt{1 - \sin^2(\beta)} = \sqrt{1 - \left(\frac{1}{2.6}\right)^2} = \sqrt{1 - \frac{1}{6.76}} = \sqrt{\frac{5.76}{6.76}} = \frac{2.4}{2.6} \]
Hence,
\[ \cot(\beta) = \frac{\cos(\beta)}{\sin(\beta)} = \frac{\frac{2.4}{2.6}}{\frac{1}{2.6}} = 2.4 \]
Therefore,
\[ OB = h \cdot 2.4 \]

### Step 3: Apply the Pythagorean Theorem
Since O is the midpoint of BC, we can apply the Pythagorean theorem in triangle AOB:
\[ AB^2 = OA^2 + OB^2 \]
Substituting the values we have:
\[ 100^2 = (h \cdot 3.2)^2 + (h \cdot 2.4)^2 \]
This simplifies to:
\[ 10000 = (10.24h^2) + (5.76h^2) \]
Combining the terms:
\[ 10000 = 16h^2 \]

### Step 4: Solve for \( h \)
Now, we can solve for \( h \):
\[ h^2 = \frac{10000}{16} = 625 \]
Taking the square root:
\[ h = \sqrt{625} = 25 \, \text{m} \]

### Final Answer
The height of the clock tower is \( 25 \, \text{m} \).

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