The number of values of b for which there is an isoscles triangle with sides of length `b + 5,3b - 2`, and `6-b` is:

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the number of values of \( b \) for which there is an isosceles triangle with sides of length \( b + 5 \), \( 3b - 2 \), and \( 6 - b \), we will set up equations based on the property of isosceles triangles, which states that at least two sides must be equal.

### Step 1: Set up the equations for the isosceles triangle

We will consider three cases where two sides are equal:

1. **Case 1:** \( b + 5 = 3b - 2 \)
2. **Case 2:** \( 3b - 2 = 6 - b \)
3. **Case 3:** \( b + 5 = 6 - b \)

### Step 2: Solve Case 1

For Case 1:
\[ b + 5 = 3b - 2 \]
Rearranging gives:
\[ 5 + 2 = 3b - b \]
\[ 7 = 2b \]
\[ b = \frac{7}{2} \]

### Step 3: Check the sides for Case 1

Now we check the sides:
- \( b + 5 = \frac{7}{2} + 5 = \frac{7}{2} + \frac{10}{2} = \frac{17}{2} \)
- \( 3b - 2 = 3 \times \frac{7}{2} - 2 = \frac{21}{2} - \frac{4}{2} = \frac{17}{2} \)
- \( 6 - b = 6 - \frac{7}{2} = \frac{12}{2} - \frac{7}{2} = \frac{5}{2} \)

The sides are \( \frac{17}{2}, \frac{17}{2}, \frac{5}{2} \). This case is valid.

### Step 4: Solve Case 2

For Case 2:
\[ 3b - 2 = 6 - b \]
Rearranging gives:
\[ 3b + b = 6 + 2 \]
\[ 4b = 8 \]
\[ b = 2 \]

### Step 5: Check the sides for Case 2

Now we check the sides:
- \( b + 5 = 2 + 5 = 7 \)
- \( 3b - 2 = 3 \times 2 - 2 = 6 - 2 = 4 \)
- \( 6 - b = 6 - 2 = 4 \)

The sides are \( 7, 4, 4 \). This case is valid.

### Step 6: Solve Case 3

For Case 3:
\[ b + 5 = 6 - b \]
Rearranging gives:
\[ b + b = 6 - 5 \]
\[ 2b = 1 \]
\[ b = \frac{1}{2} \]

### Step 7: Check the sides for Case 3

Now we check the sides:
- \( b + 5 = \frac{1}{2} + 5 = \frac{1}{2} + \frac{10}{2} = \frac{11}{2} \)
- \( 3b - 2 = 3 \times \frac{1}{2} - 2 = \frac{3}{2} - 2 = \frac{3}{2} - \frac{4}{2} = -\frac{1}{2} \) (not valid)
- \( 6 - b = 6 - \frac{1}{2} = \frac{12}{2} - \frac{1}{2} = \frac{11}{2} \)

The sides are \( \frac{11}{2}, -\frac{1}{2}, \frac{11}{2} \). Since one side is negative, this case is invalid.

### Conclusion

The valid values of \( b \) are from Case 1 and Case 2, which are:
1. \( b = \frac{7}{2} \)
2. \( b = 2 \)

Thus, the number of values of \( b \) for which there is an isosceles triangle is **2**.

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