If `p_(1), p_(2),p_(3)` are respectively the perpendiculars from the vertices of a triangle to the opposite sides , then `(cosA)/(p_(1))+(cosB)/(p_(2))+(cosC)/(p_(3))` is equal to

Options

A. `1//r`
B. `1//R`
C. `1//!`
D. none of these

Answer

To solve the problem, we need to find the value of the expression:

\[ \frac{\cos A}{p_1} + \frac{\cos B}{p_2} + \frac{\cos C}{p_3} \]

where \( p_1, p_2, p_3 \) are the perpendiculars from the vertices of triangle \( ABC \) to the opposite sides.

### Step 1: Understanding the Perpendiculars
The perpendiculars \( p_1, p_2, p_3 \) can be expressed in terms of the area \( \Delta \) of triangle \( ABC \) and the sides \( a, b, c \) opposite to angles \( A, B, C \) respectively:

\[ p_1 = \frac{2\Delta}{a}, \quad p_2 = \frac{2\Delta}{b}, \quad p_3 = \frac{2\Delta}{c} \]

### Step 2: Substitute the Perpendiculars
Now, substituting these expressions into our original equation:

\[ \frac{\cos A}{p_1} = \frac{\cos A \cdot a}{2\Delta}, \quad \frac{\cos B}{p_2} = \frac{\cos B \cdot b}{2\Delta}, \quad \frac{\cos C}{p_3} = \frac{\cos C \cdot c}{2\Delta} \]

Thus, we can rewrite the expression as:

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

### Step 3: Combine the Terms
Combining these terms gives:

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

### Step 4: Use the Sine Rule
Using the sine rule, we know that:

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]

where \( R \) is the circumradius of triangle \( ABC \). Therefore, we can express \( a, b, c \) in terms of \( R \) and the sines of the angles:

\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]

### Step 5: Substitute Back into the Expression
Substituting these back into our expression:

\[ a \cos A + b \cos B + c \cos C = 2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C \]

Using the identity \( \sin A \cos A = \frac{1}{2} \sin 2A \):

\[ = 2R \left( \frac{1}{2} \sin 2A + \frac{1}{2} \sin 2B + \frac{1}{2} \sin 2C \right) = R (\sin 2A + \sin 2B + \sin 2C) \]

### Step 6: Final Expression
Now substituting this back into our earlier expression:

\[ \frac{1}{2\Delta} \cdot R (\sin 2A + \sin 2B + \sin 2C) \]

### Step 7: Area of Triangle
The area \( \Delta \) can also be expressed in terms of \( R \):

\[ \Delta = \frac{abc}{4R} \]

Thus, we can simplify our expression further:

\[ \frac{R (\sin 2A + \sin 2B + \sin 2C)}{2 \cdot \frac{abc}{4R}} = \frac{2R^2 (\sin 2A + \sin 2B + \sin 2C)}{abc} \]

### Conclusion
After simplifying, we find that:

\[ \frac{\cos A}{p_1} + \frac{\cos B}{p_2} + \frac{\cos C}{p_3} = \frac{1}{R} \]

Thus, the final answer is:

\[ \frac{1}{R} \]