Let triangle ABC is right triangle right angled at C such that `A lt B` and `r=8, R=41` .
Q. `"tan"(A)/(2)=`

Options

A. `(1)/(18)`
B. `(1)/(3)`
C. `(1)/(6)`
D. `(1)/(9)`

Answer

To solve the problem, we need to find the value of \( \frac{\tan(A)}{2} \) for triangle \( ABC \) which is a right triangle with \( C \) as the right angle. Given that the inradius \( r = 8 \) and the circumradius \( R = 41 \), we can use the relationships between the sides and angles of the triangle.

### Step-by-Step Solution:

1. **Understanding the Triangle Properties**:
- In a right triangle, the circumradius \( R \) is given by the formula:
\[ R = \frac{c}{2} \]
where \( c \) is the hypotenuse. Therefore, we have:
\[ c = 2R = 2 \times 41 = 82 \]

2. **Using the Inradius**:
- The inradius \( r \) is given by the formula:
\[ r = \frac{A}{s} \]
where \( A \) is the area of the triangle and \( s \) is the semi-perimeter. The semi-perimeter \( s \) can also be expressed as:
\[ s = \frac{a + b + c}{2} \]
where \( a \) and \( b \) are the legs of the triangle.

3. **Finding the Semi-Perimeter**:
- Since \( r = 8 \):
\[ A = r \cdot s \]
- The area \( A \) of triangle \( ABC \) can also be expressed as:
\[ A = \frac{1}{2}ab \]
- Therefore, we have:
\[ 8s = \frac{1}{2}ab \]
- Substituting \( s \):
\[ 8 \cdot \frac{a + b + 82}{2} = \frac{1}{2}ab \]
- Simplifying this gives:
\[ 8(a + b + 82) = ab \]

4. **Using Pythagorean Theorem**:
- From the Pythagorean theorem, we know:
\[ a^2 + b^2 = c^2 = 82^2 = 6724 \]

5. **Setting Up the Equations**:
- We now have two equations:
1. \( ab = 8(a + b + 82) \)
2. \( a^2 + b^2 = 6724 \)

6. **Solving the System of Equations**:
- Let \( s = a + b \) and \( p = ab \). From the first equation, we can express \( p \):
\[ p = 8(s + 82) \]
- We can also express \( a^2 + b^2 \) in terms of \( s \) and \( p \):
\[ a^2 + b^2 = s^2 - 2p \]
- Substituting \( p \):
\[ s^2 - 2(8(s + 82)) = 6724 \]
- Simplifying this gives:
\[ s^2 - 16s - 164 = 6724 \]
- Rearranging:
\[ s^2 - 16s - 6888 = 0 \]

7. **Finding the Roots**:
- Using the quadratic formula \( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ s = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot (-6888)}}{2 \cdot 1} \]
\[ s = \frac{16 \pm \sqrt{256 + 27552}}{2} \]
\[ s = \frac{16 \pm \sqrt{27708}}{2} \]

8. **Calculating \( \tan(A) \)**:
- Once we find \( a \) and \( b \), we can find \( \tan(A) \) using:
\[ \tan(A) = \frac{a}{b} \]
- Finally, we need \( \frac{\tan(A)}{2} \).

### Final Calculation:
After calculating the values of \( a \) and \( b \), we can substitute them back to find \( \tan(A) \) and then divide by 2 to get \( \frac{\tan(A)}{2} \).