Let `cosA+cosB + cos C=3/2` in a triangle then the type of the triangle is

Answer

To determine the type of triangle given that \( \cos A + \cos B + \cos C = \frac{3}{2} \), we can follow these steps:

### Step 1: Use the identity for cosines
We know that in any triangle, the angles \( A \), \( B \), and \( C \) satisfy \( A + B + C = 180^\circ \). We can express \( C \) in terms of \( A \) and \( B \):
\[ C = 180^\circ - A - B \]

### Step 2: Substitute \( C \) into the cosine equation
Substituting \( C \) into the equation gives us:
\[ \cos A + \cos B + \cos(180^\circ - A - B) = \frac{3}{2} \]
Using the property \( \cos(180^\circ - x) = -\cos x \), we have:
\[ \cos A + \cos B - \cos(A + B) = \frac{3}{2} \]

### Step 3: Use the cosine addition formula
Using the cosine addition formula:
\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]
We can rewrite the equation as:
\[ \cos A + \cos B - (\cos A \cos B - \sin A \sin B) = \frac{3}{2} \]
This simplifies to:
\[ \cos A + \cos B + \sin A \sin B - \cos A \cos B = \frac{3}{2} \]

### Step 4: Analyze the equation
To analyze the equation, we can consider the maximum values of \( \cos A \) and \( \cos B \). Since \( \cos A \) and \( \cos B \) can each be at most 1, the maximum value of \( \cos A + \cos B \) is 2. Thus, we can deduce that:
\[ \cos A + \cos B \leq 2 \]
This implies that the maximum value of \( \cos A + \cos B + \cos C \) is also limited, and if \( \cos A + \cos B + \cos C = \frac{3}{2} \), this suggests that the angles \( A \), \( B \), and \( C \) must be such that they are all less than \( 90^\circ \).

### Step 5: Conclude the type of triangle
If \( A \), \( B \), and \( C \) are all less than \( 90^\circ \) and equal (since \( \cos A + \cos B + \cos C = \frac{3}{2} \) suggests symmetry), we conclude that:
\[ A = B = C = 60^\circ \]
Thus, the triangle is equilateral.

### Final Answer
The type of triangle is **equilateral**.

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