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

Options

A. `(b sin 2 alpha)/(2(1 -cos alpha))`
B. `(b sin 2 alpha)/(2(1 + cos alpha))`
C. `(b sin alpha)/(2)`
D. `(b sin alpha)/(2(1+ sin alpha))`

Explanation

To find the inradius \( r \) of an isosceles triangle with equal sides of length \( b \) and angle \( \alpha < \frac{\pi}{4} \), we can follow these steps:

### Step-by-Step Solution:

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

2. **Drop a Perpendicular**:
Let \( AD \) be the perpendicular from vertex \( A \) to the base \( BC \). Since \( ABC \) is isosceles, \( D \) is the midpoint of \( BC \).

3. **Calculate Lengths**:
The length of \( BD \) (half of base \( BC \)) can be calculated using the sine function:
\[ BD = b \sin(\alpha) \]
The length of \( AD \) (the height) can be calculated using the cosine function:
\[ AD = b \cos(\alpha) \]

4. **Calculate the Area \( \Delta \)**:
The area \( \Delta \) of triangle \( ABC \) can be calculated as:
\[ \Delta = \frac{1}{2} \times BC \times AD \]
The length of \( BC \) can be calculated as:
\[ BC = 2 \times BD = 2b \sin(\alpha) \]
Thus, the area becomes:
\[ \Delta = \frac{1}{2} \times (2b \sin(\alpha)) \times (b \cos(\alpha)) = b^2 \sin(\alpha) \cos(\alpha) \]

5. **Calculate the Semi-Perimeter \( s \)**:
The semi-perimeter \( s \) of triangle \( ABC \) is given by:
\[ s = \frac{AB + AC + BC}{2} = \frac{b + b + 2b \sin(\alpha)}{2} = b \left(1 + \sin(\alpha)\right) \]

6. **Calculate the Inradius \( r \)**:
The inradius \( r \) can be calculated using the formula:
\[ r = \frac{\Delta}{s} \]
Substituting the values we found:
\[ r = \frac{b^2 \sin(\alpha) \cos(\alpha)}{b(1 + \sin(\alpha))} \]
Simplifying this gives:
\[ r = \frac{b \sin(\alpha) \cos(\alpha)}{1 + \sin(\alpha)} \]

7. **Further Simplification**:
Using the identity \( \sin(2\alpha) = 2 \sin(\alpha) \cos(\alpha) \), we can express \( r \) as:
\[ r = \frac{b}{2(1 + \sin(\alpha))} \sin(2\alpha) \]

### Final Result:
Thus, the inradius \( r \) of the isosceles triangle is given by:
\[ r = \frac{b \sin(2\alpha)}{2(1 + \sin(\alpha))} \]