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 perimeters of `DeltaLMN and Delta ABC an lamda and mu, ` then the value of `(lamda)/(mu)` is

Options

A. `r/R`
B. `R/r`
C. `(rR)/(Delta)`
D. `(Delta)/(rR)`

Answer

To solve the problem, we need to find the ratio \(\frac{\lambda}{\mu}\), where \(\lambda\) is the perimeter of triangle \(LMN\) and \(\mu\) is the perimeter of triangle \(ABC\).

### Step-by-Step Solution:

1. **Understanding the Perimeters**:
- The perimeter of triangle \(ABC\) is denoted as \(\mu\).
- The perimeter of triangle \(LMN\) is denoted as \(\lambda\).

2. **Finding the Perimeter of Triangle \(ABC\)**:
- The perimeter \(\mu\) of triangle \(ABC\) can be expressed as:
\[ \mu = AB + BC + CA = a + b + c \]
where \(a\), \(b\), and \(c\) are the lengths of the sides of triangle \(ABC\).

3. **Finding the Perimeter of Triangle \(LMN\)**:
- The sides of triangle \(LMN\) can be expressed in terms of the angles and sides of triangle \(ABC\). The lengths of the sides of triangle \(LMN\) can be given by:
\[ LM = BC \cdot \cos A, \quad MN = CA \cdot \cos B, \quad NL = AB \cdot \cos C \]
- Thus, the perimeter \(\lambda\) can be expressed as:
\[ \lambda = LM + MN + NL = (BC \cdot \cos A) + (CA \cdot \cos B) + (AB \cdot \cos C \]

4. **Using the Sine Rule**:
- We can relate the sides of triangle \(ABC\) to its circumradius \(R\) and area \(\Delta\):
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]

5. **Expressing \(\lambda\) in Terms of \(R\)**:
- Substituting the expressions for \(a\), \(b\), and \(c\) into the perimeter of triangle \(LMN\):
\[ \lambda = 2R(\sin A \cdot \cos A + \sin B \cdot \cos B + \sin C \cdot \cos C) \]
- Using the identity \(\sin A \cos A = \frac{1}{2} \sin 2A\), we can rewrite \(\lambda\):
\[ \lambda = R(\sin 2A + \sin 2B + \sin 2C) \]

6. **Finding the Ratio \(\frac{\lambda}{\mu}\)**:
- We know that the perimeter \(\mu\) can also be expressed in terms of the semi-perimeter \(s\):
\[ \mu = 2s = 2 \cdot \frac{a + b + c}{2} = a + b + c \]
- Using the relationship between area \(\Delta\), inradius \(r\), and semi-perimeter \(s\):
\[ \Delta = r \cdot s \]
- Thus, we can express \(\mu\) in terms of \(r\):
\[ \mu = 2s = \frac{2\Delta}{r} \]

7. **Final Ratio**:
- Now we can find the ratio:
\[ \frac{\lambda}{\mu} = \frac{R(\sin 2A + \sin 2B + \sin 2C)}{\frac{2\Delta}{r}} = \frac{R \cdot r}{\Delta} \]
- After simplification, we find that:
\[ \frac{\lambda}{\mu} = \frac{r}{R} \]

### Conclusion:
The value of \(\frac{\lambda}{\mu}\) is \(\frac{r}{R}\).