AL, BM and CN are perpendicular from angular points of a triangle ABC on the opposite sides BC, CA and AB respectively. `Delta` is the area of triangle ABC, (r ) and R are the inradius and circumradius.
If area of `Delta LMN` is `Delta ',` then the value of `(Delta')/(Delta)` is

Options

A. `2 sin A sin B sin C`
B. `2 cos A cos B cos C`
C. `sin A sin B sin C`
D. `cos A cos B cos C`

Answer

To solve the problem, we need to find the ratio of the area of triangle \( \Delta' \) (formed by the feet of the altitudes from the vertices of triangle \( ABC \) to the opposite sides) to the area of triangle \( ABC \) (denoted as \( \Delta \)).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let \( A, B, C \) be the vertices of triangle \( ABC \).
- Let \( L, M, N \) be the feet of the perpendiculars from \( A, B, C \) to the sides \( BC, CA, AB \) respectively.

2. **Area of Triangle \( ABC \)**:
- The area \( \Delta \) of triangle \( ABC \) can be expressed using the formula:
\[ \Delta = \frac{1}{2} \times BC \times h_A \]
where \( h_A \) is the altitude from vertex \( A \) to side \( BC \).

3. **Finding Areas of Smaller Triangles**:
- The area of triangle \( AMN \) can be calculated as:
\[ \Delta_1 = \frac{1}{2} \times AM \times AN \times \sin A \]
- From the properties of triangles, we have:
\[ AM = BC \cdot \cos A \quad \text{and} \quad AN = AC \cdot \cos A \]
- Thus, substituting these into the area formula gives:
\[ \Delta_1 = \frac{1}{2} \times (BC \cdot \cos A) \times (AC \cdot \cos A) \times \sin A = \frac{1}{2} \cdot BC \cdot AC \cdot \sin A \cdot \cos^2 A \]
- This can be expressed in terms of \( \Delta \):
\[ \Delta_1 = \Delta \cdot \cos^2 A \]

4. **Calculating Areas of Other Triangles**:
- Similarly, for triangles \( BLM \) and \( CLN \):
\[ \Delta_2 = \Delta \cdot \cos^2 B \]
\[ \Delta_3 = \Delta \cdot \cos^2 C \]

5. **Total Area of Triangle \( LMN \)**:
- The total area \( \Delta' \) of triangle \( LMN \) is:
\[ \Delta' = \Delta - (\Delta_1 + \Delta_2 + \Delta_3) = \Delta - \Delta(\cos^2 A + \cos^2 B + \cos^2 C) \]
- This simplifies to:
\[ \Delta' = \Delta \left(1 - (\cos^2 A + \cos^2 B + \cos^2 C)\right) \]

6. **Using Trigonometric Identity**:
- We can use the identity \( \cos^2 A + \cos^2 B + \cos^2 C = 1 - 2\cos A \cos B \cos C \):
\[ \Delta' = \Delta \cdot (1 - (1 - 2\cos A \cos B \cos C)) = \Delta \cdot (2\cos A \cos B \cos C) \]

7. **Finding the Ratio**:
- Finally, the ratio \( \frac{\Delta'}{\Delta} \) is:
\[ \frac{\Delta'}{\Delta} = 2 \cos A \cos B \cos C \]

### Final Answer:
\[ \frac{\Delta'}{\Delta} = 2 \cos A \cos B \cos C \]