In a `DeltaABC, cos A+cos B+cosC` belongs to the interval

Options

A. `(1//2,3//2)`
B. `(1//,3//2)`
C. `(3//2,2)`
D. none of these

Explanation

To solve the problem of finding the interval for \( \cos A + \cos B + \cos C \) in triangle \( ABC \), we can follow these steps:

### Step 1: Use the Cosine Sum Identity
We start with the expression \( \cos A + \cos B + \cos C \). We can use the cosine sum identity to rewrite \( \cos A + \cos B \):
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Since \( A + B + C = 180^\circ \), we can express \( C \) as \( C = 180^\circ - (A + B) \).

### Step 2: Substitute for \( C \)
Substituting \( C \) into the expression, we have:
\[ \cos C = \cos(180^\circ - (A + B)) = -\cos(A + B) \]
Thus, we can rewrite \( \cos A + \cos B + \cos C \) as:
\[ \cos A + \cos B - \cos(A + B) \]

### Step 3: Analyze the Range
To analyze the range of \( \cos A + \cos B + \cos C \), we know that the values of \( A \), \( B \), and \( C \) are constrained by the triangle inequality and the fact that they must sum to \( 180^\circ \).

### Step 4: Maximum and Minimum Values
1. **Maximum Value**: The maximum occurs when \( A = B = C = 60^\circ \):
\[ \cos 60^\circ + \cos 60^\circ + \cos 60^\circ = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \]

2. **Minimum Value**: The minimum occurs when one angle approaches \( 0^\circ \) and the other two angles approach \( 90^\circ \):
\[ \cos 0^\circ + \cos 90^\circ + \cos 90^\circ = 1 + 0 + 0 = 1 \]

### Step 5: Conclusion
Thus, we conclude that:
\[ 1 \leq \cos A + \cos B + \cos C \leq \frac{3}{2} \]
Therefore, the interval for \( \cos A + \cos B + \cos C \) is:
\[ \boxed{[1, \frac{3}{2}]} \]