The area of an acute triangle ABC is `Delta`, the area of its pedal triangle is 'p' , where `cos B=(2p)/(Delta)` and `sin B=(2sqrt(3)p)/(Delta)`. The value of `8(cos^(2)A cos B+cos^(2)C)` is

Options

A. 1
B. 2
C. 3
D. none of these

Explanation

To solve the problem, we need to find the value of \( 8(\cos^2 A \cos B + \cos^2 C) \) given the conditions about the triangle ABC.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We have the following:
- Area of triangle ABC: \( \Delta \)
- Area of the pedal triangle: \( p \)
- \( \cos B = \frac{2p}{\Delta} \)
- \( \sin B = \frac{2\sqrt{3}p}{\Delta} \)

2. **Using the Pythagorean Identity**:
From the identity \( \cos^2 B + \sin^2 B = 1 \):
\[ \left(\frac{2p}{\Delta}\right)^2 + \left(\frac{2\sqrt{3}p}{\Delta}\right)^2 = 1 \]
Simplifying this:
\[ \frac{4p^2}{\Delta^2} + \frac{12p^2}{\Delta^2} = 1 \]
\[ \frac{16p^2}{\Delta^2} = 1 \]
\[ 16p^2 = \Delta^2 \implies \Delta = 4p \]

3. **Finding \( \frac{p}{\Delta} \)**:
\[ \frac{p}{\Delta} = \frac{p}{4p} = \frac{1}{4} \]

4. **Using the Relationship of Areas**:
We have established that \( \Delta = 4p \). Now we can express \( \cos B \) and \( \sin B \) in terms of \( p \):
\[ \cos B = \frac{2p}{4p} = \frac{1}{2} \]
\[ \sin B = \frac{2\sqrt{3}p}{4p} = \frac{\sqrt{3}}{2} \]

5. **Identifying Angles**:
Since \( \cos B = \frac{1}{2} \) and \( \sin B = \frac{\sqrt{3}}{2} \), we can conclude that:
\[ B = 60^\circ \]

6. **Equilateral Triangle**:
Given that \( B = 60^\circ \) and the triangle is acute, we can deduce that \( A = 60^\circ \) and \( C = 60^\circ \) as well. Thus, triangle ABC is equilateral.

7. **Calculating \( 8(\cos^2 A \cos B + \cos^2 C) \)**:
Since \( A = B = C = 60^\circ \):
\[ \cos A = \cos B = \cos C = \frac{1}{2} \]
Therefore:
\[ \cos^2 A = \cos^2 B = \cos^2 C = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
Now substituting into the expression:
\[ 8\left(\cos^2 A \cos B + \cos^2 C\right) = 8\left(\frac{1}{4} \cdot \frac{1}{2} + \frac{1}{4}\right) \]
\[ = 8\left(\frac{1}{8} + \frac{1}{4}\right) = 8\left(\frac{1}{8} + \frac{2}{8}\right) = 8\left(\frac{3}{8}\right) = 3 \]

### Final Answer:
The value of \( 8(\cos^2 A \cos B + \cos^2 C) \) is \( 3 \).