In triangle ABC, `b^(2) sin 2C + c^(2) sin 2B = 2bc` where `b = 20, c = 21`, then inradius =

Options

A. 4
B. 6
C. 8
D. 9

Explanation

To solve the problem, we need to find the inradius of triangle ABC given the equation \( b^2 \sin 2C + c^2 \sin 2B = 2bc \) with \( b = 20 \) and \( c = 21 \).

### Step-by-Step Solution:

1. **Write down the given equation:**
\[ b^2 \sin 2C + c^2 \sin 2B = 2bc \]

2. **Substitute the values of \( b \) and \( c \):**
\[ 20^2 \sin 2C + 21^2 \sin 2B = 2 \cdot 20 \cdot 21 \]
Simplifying gives:
\[ 400 \sin 2C + 441 \sin 2B = 840 \]

3. **Use the double angle formula for sine:**
\[ \sin 2C = 2 \sin C \cos C \quad \text{and} \quad \sin 2B = 2 \sin B \cos B \]
Substitute these into the equation:
\[ 400 \cdot 2 \sin C \cos C + 441 \cdot 2 \sin B \cos B = 840 \]
This simplifies to:
\[ 800 \sin C \cos C + 882 \sin B \cos B = 840 \]

4. **Multiply the entire equation by \( 2r \):**
\[ 800 \cdot 2r \sin C \cos C + 882 \cdot 2r \sin B \cos B = 840 \cdot 2r \]

5. **Use the sine rule:**
From the sine rule, we know:
\[ a = 2r \sin A, \quad b = 2r \sin B, \quad c = 2r \sin C \]
Substitute these into the equation:
\[ b^2 \cdot 2r \sin C + c^2 \cdot 2r \sin B = 2bc \]

6. **Rearranging gives:**
\[ b \cos C + c \cos B = 2r \]

7. **Using the projection formula:**
From the projection formula, we have:
\[ a = b \cos C + c \cos B \]
Therefore, we can conclude:
\[ a = 2r \]

8. **Since \( a = 2r \), we can find \( a \):**
To find \( a \), we use the Pythagorean theorem since we suspect that triangle ABC is a right triangle:
\[ a^2 = b^2 + c^2 \]
Substituting the values:
\[ a^2 = 20^2 + 21^2 = 400 + 441 = 841 \]
Therefore:
\[ a = \sqrt{841} = 29 \]

9. **Calculate the semi-perimeter \( s \):**
\[ s = \frac{a + b + c}{2} = \frac{29 + 20 + 21}{2} = 35 \]

10. **Calculate the area \( \Delta \):**
The area \( \Delta \) of triangle ABC can be calculated using:
\[ \Delta = \frac{1}{2} \times b \times c = \frac{1}{2} \times 20 \times 21 = 210 \]

11. **Calculate the inradius \( r \):**
The inradius \( r \) is given by:
\[ r = \frac{\Delta}{s} = \frac{210}{35} = 6 \]

### Final Answer:
The inradius \( r \) of triangle ABC is \( 6 \).