In triangle ABC, if `cos A + cos B + cos C = (7)/(4), " then " (R)/(r)` is equal to

Options

A. `(3)/(4)`
B. `(4)/(3)`
C. `(2)/(3)`
D. `(3)/(2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given condition
We are given that in triangle ABC,
\[ \cos A + \cos B + \cos C = \frac{7}{4}. \]

### Step 2: Use the identity for cosines
We know from trigonometric identities that:
\[ \cos A + \cos B + \cos C = 1 + 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}. \]
Setting this equal to the given value, we have:
\[ 1 + 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{7}{4}. \]

### Step 3: Rearrange the equation
Subtract 1 from both sides:
\[ 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{7}{4} - 1. \]
Calculating the right side:
\[ \frac{7}{4} - 1 = \frac{7}{4} - \frac{4}{4} = \frac{3}{4}. \]
Thus, we have:
\[ 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{3}{4}. \]

### Step 4: Simplify the equation
Dividing both sides by 4 gives:
\[ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{3}{16}. \]

### Step 5: Relate inradius (r) and circumradius (R)
We know the formula relating the inradius \( r \) and circumradius \( R \) of a triangle:
\[ r = R \cdot 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}. \]
Substituting the value we found:
\[ r = R \cdot 4 \cdot \frac{3}{16} = R \cdot \frac{3}{4}. \]

### Step 6: Find the ratio \( \frac{R}{r} \)
Rearranging gives us:
\[ \frac{R}{r} = \frac{R}{R \cdot \frac{3}{4}} = \frac{4}{3}. \]

### Conclusion
Thus, we find that:
\[ \frac{R}{r} = \frac{4}{3}. \]

### Final Answer
The value of \( \frac{R}{r} \) is \( \frac{4}{3} \).

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