Which of the following is possible in `DeltaABC`?

Options

A. `cosA+cosB+cosC=3/2`
B. `cosAcosBcosC=0`
C. `sinA+sinB+sinC=sqrt2+1`
D. `sinAsinBsinC=-3/8`

Explanation

To determine which of the options is possible in triangle \( \Delta ABC \), we will analyze each option step by step.

### Step 1: Analyze Option A
**Option A:** \( \cos A + \cos B + \cos C = \frac{3}{2} \)

For a triangle, the angles \( A, B, C \) must satisfy \( A + B + C = 180^\circ \) or \( A + B + C = \pi \) radians.

In an equilateral triangle, where \( A = B = C = 60^\circ \):
\[ \cos 60^\circ = \frac{1}{2} \]
Thus,
\[ \cos A + \cos B + \cos C = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \]
This option is possible.

### Step 2: Analyze Option B
**Option B:** \( \cos A \cdot \cos B \cdot \cos C = 0 \)

This condition is satisfied if at least one of the angles \( A, B, \) or \( C \) is \( 90^\circ \) (i.e., the triangle is a right triangle). If \( A = 90^\circ \):
\[ \cos A = 0 \]
Thus,
\[ \cos A \cdot \cos B \cdot \cos C = 0 \]
This option is possible.

### Step 3: Analyze Option C
**Option C:** \( \sin A + \sin B + \sin C = \sqrt{2} + 1 \)

For this to hold, we can consider the case where \( A = 90^\circ \) and \( B = C = 45^\circ \):
\[ \sin 90^\circ = 1, \quad \sin 45^\circ = \frac{1}{\sqrt{2}} \]
Thus,
\[ \sin A + \sin B + \sin C = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = 1 + \frac{2}{\sqrt{2}} = 1 + \sqrt{2} \]
This option is also possible.

### Step 4: Analyze Option D
**Option D:** \( \sin A \cdot \sin B \cdot \sin C = -\frac{3}{8} \)

Since the sine of any angle in a triangle is always non-negative (as angles \( A, B, C \) are between \( 0^\circ \) and \( 180^\circ \)), the product \( \sin A \cdot \sin B \cdot \sin C \) cannot be negative. Thus, this option is not possible.

### Conclusion
The possible options in triangle \( \Delta ABC \) are:
- Option A: Correct
- Option B: Correct
- Option C: Correct
- Option D: Incorrect

### Final Answer
The correct options are A, B, and C.

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