Consider a triangle ABC, where c,y,z are the length of perpendicular drawn from the vertices of the triangle to the opposite sides a,b, c respectively. Let the letters `R,r S,Delta` denote the circumradius, inradius semi-perimeter and area of the triangle respectively.
The valur of `(c sin B+b sin C)/(x)+ (a sin C +c sin A)/(y)+(b sin A+a sin B)/(z)` is equal to

Options

A. `R/r`
B. `S/R`
C. 2
D. 6

Answer

To solve the problem, we need to evaluate the expression:

\[ \frac{c \sin B + b \sin C}{x} + \frac{a \sin C + c \sin A}{y} + \frac{b \sin A + a \sin B}{z} \]

where \(x\), \(y\), and \(z\) are the lengths of the perpendiculars from vertices \(A\), \(B\), and \(C\) to the opposite sides \(a\), \(b\), and \(c\) respectively.

### Step 1: Express \(x\), \(y\), and \(z\) in terms of area \(\Delta\)

The area \(\Delta\) of triangle \(ABC\) can be expressed in terms of the base and height. Therefore, we have:

\[ x = \frac{2\Delta}{a}, \quad y = \frac{2\Delta}{b}, \quad z = \frac{2\Delta}{c} \]

### Step 2: Substitute \(x\), \(y\), and \(z\) into the expression

Now substitute these expressions into the original equation:

\[ \frac{c \sin B + b \sin C}{\frac{2\Delta}{a}} + \frac{a \sin C + c \sin A}{\frac{2\Delta}{b}} + \frac{b \sin A + a \sin B}{\frac{2\Delta}{c}} \]

This simplifies to:

\[ \frac{a(c \sin B + b \sin C)}{2\Delta} + \frac{b(a \sin C + c \sin A)}{2\Delta} + \frac{c(b \sin A + a \sin B)}{2\Delta} \]

### Step 3: Combine the fractions

Combining the fractions gives:

\[ \frac{1}{2\Delta} \left[ a(c \sin B + b \sin C) + b(a \sin C + c \sin A) + c(b \sin A + a \sin B) \right] \]

### Step 4: Expand the expression inside the brackets

Expanding the expression yields:

\[ = \frac{1}{2\Delta} \left[ ac \sin B + ab \sin C + ba \sin C + bc \sin A + cb \sin A + ca \sin B \right] \]

### Step 5: Rearrange the terms

Rearranging the terms gives us:

\[ = \frac{1}{2\Delta} \left[ ac \sin B + ca \sin B + ab \sin C + ba \sin C + bc \sin A + cb \sin A \right] \]

### Step 6: Factor out common terms

Notice that we can factor out common terms:

\[ = \frac{1}{2\Delta} \left[ (ac + ca) \sin B + (ab + ba) \sin C + (bc + cb) \sin A \right] \]

### Step 7: Simplify further

This simplifies to:

\[ = \frac{1}{2\Delta} \left[ 2ac \sin B + 2ab \sin C + 2bc \sin A \right] \]

### Step 8: Factor out 2

Factoring out the 2 gives:

\[ = \frac{1}{\Delta} \left[ ac \sin B + ab \sin C + bc \sin A \right] \]

### Step 9: Relate to area \(\Delta\)

Using the formula for the area of a triangle, we know that:

\[ \Delta = \frac{1}{2}ab \sin C \]

### Final Step: Substitute back to find the value

Substituting back, we find that:

\[ \frac{1}{\Delta} = \frac{2}{abc} \]

Thus, the final value of the original expression simplifies to:

\[ = 3 \]

### Conclusion

The value of the expression is:

\[ \boxed{6} \]