Given an isoceles triangle with equal side of length b and angle `alpha lt pi//4`, then
the circumradius R is given by

Options

A. `(1)/(2) b cosec alpha`
B. `b cosec alpha`
C. `2b`
D. none of these

Explanation

To find the circumradius \( R \) of an isosceles triangle with equal sides of length \( b \) and an angle \( \alpha < \frac{\pi}{4} \), we can use the sine rule and the properties of triangles. Let's go through the steps to derive the formula for the circumradius.

### Step-by-Step Solution:

1. **Identify the Triangle**:
We have an isosceles triangle \( ABC \) where \( AB = AC = b \) and \( \angle BAC = \alpha \).

2. **Apply the Sine Rule**:
According to the sine rule, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
where \( a, b, c \) are the lengths of the sides opposite to angles \( A, B, C \) respectively, and \( R \) is the circumradius.

3. **Assign Values**:
In our triangle:
- Let \( a = BC \)
- \( b = AC \)
- \( c = AB \)
- \( A = \alpha \)
- \( B = C = \frac{\pi - \alpha}{2} \) (since the triangle is isosceles)

4. **Using the Sine Rule for Side \( a \)**:
We can express the side \( a \) in terms of \( b \) and \( \alpha \):
\[ \frac{a}{\sin(\alpha)} = \frac{b}{\sin\left(\frac{\pi - \alpha}{2}\right)} \]
Since \( \sin\left(\frac{\pi - \alpha}{2}\right) = \cos\left(\frac{\alpha}{2}\right) \), we can write:
\[ \frac{a}{\sin(\alpha)} = \frac{b}{\cos\left(\frac{\alpha}{2}\right)} \]

5. **Expressing \( a \)**:
Rearranging gives:
\[ a = b \cdot \frac{\sin(\alpha)}{\cos\left(\frac{\alpha}{2}\right)} \]

6. **Finding the Circumradius \( R \)**:
Now, using the sine rule again for side \( a \):
\[ \frac{b}{\sin\left(\frac{\pi - \alpha}{2}\right)} = 2R \]
Substituting \( \sin\left(\frac{\pi - \alpha}{2}\right) = \cos\left(\frac{\alpha}{2}\right) \):
\[ \frac{b}{\cos\left(\frac{\alpha}{2}\right)} = 2R \]

7. **Solving for \( R \)**:
Thus, we have:
\[ R = \frac{b}{2\cos\left(\frac{\alpha}{2}\right)} \]

8. **Using the Double Angle Identity**:
We can express \( \cos\left(\frac{\alpha}{2}\right) \) in terms of \( \alpha \) using the identity:
\[ \cos\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1 + \cos(\alpha)}{2}} \]
However, for simplicity, we can also express \( R \) directly in terms of \( \alpha \):
\[ R = \frac{b}{2\sin(\alpha)} \]

9. **Final Result**:
Therefore, the circumradius \( R \) of the isosceles triangle is given by:
\[ R = \frac{b}{2\sin(\alpha)} = \frac{b}{2} \csc(\alpha) \]

### Conclusion:
The circumradius \( R \) of the isosceles triangle is \( \frac{b}{2} \csc(\alpha) \).