In a `△ ABC` if a =26, b= 30 and cos C =`63/65`, then `r_(2)` =

Options

A. 84
B. 45
C. 48
D. 24

Answer

To solve the problem, we will follow these steps:

### Step 1: Use the Cosine Rule to Find Side c
We know from the cosine rule that:
\[ c^2 = a^2 + b^2 - 2ab \cos C \]
Given:
- \( a = 26 \)
- \( b = 30 \)
- \( \cos C = \frac{63}{65} \)

Substituting the values into the formula:
\[ c^2 = 26^2 + 30^2 - 2 \cdot 26 \cdot 30 \cdot \frac{63}{65} \]

Calculating \( a^2 \) and \( b^2 \):
\[ c^2 = 676 + 900 - 2 \cdot 26 \cdot 30 \cdot \frac{63}{65} \]

Calculating \( 2 \cdot 26 \cdot 30 \cdot \frac{63}{65} \):
\[ = 1560 \cdot \frac{63}{65} = \frac{98280}{65} \]

Now, calculate \( 676 + 900 \):
\[ c^2 = 1576 - \frac{98280}{65} \]

Finding a common denominator:
\[ = \frac{1576 \cdot 65}{65} - \frac{98280}{65} = \frac{102440 - 98280}{65} = \frac{1756}{65} \]

Thus,
\[ c^2 = \frac{1756}{65} \]

Taking the square root:
\[ c = \sqrt{\frac{1756}{65}} = \frac{8}{\sqrt{65}} \text{ (after simplifying)} \]

### Step 2: Calculate the Semi-perimeter s
The semi-perimeter \( s \) is given by:
\[ s = \frac{a + b + c}{2} \]
Substituting the values:
\[ s = \frac{26 + 30 + 8}{2} = \frac{64}{2} = 32 \]

### Step 3: Calculate the Area (Δ) of Triangle ABC
Using Heron's formula:
\[ \Delta = \sqrt{s(s-a)(s-b)(s-c)} \]
Calculating each term:
- \( s - a = 32 - 26 = 6 \)
- \( s - b = 32 - 30 = 2 \)
- \( s - c = 32 - 8 = 24 \)

Now substituting into Heron's formula:
\[ \Delta = \sqrt{32 \cdot 6 \cdot 2 \cdot 24} \]

Calculating:
\[ = \sqrt{32 \cdot 6 \cdot 2 \cdot 24} = \sqrt{32 \cdot 288} = \sqrt{9216} = 96 \]

### Step 4: Calculate \( r_2 \)
Using the formula for \( r_2 \):
\[ r_2 = \frac{\Delta}{s - b} \]
Substituting the values:
\[ r_2 = \frac{96}{32 - 30} = \frac{96}{2} = 48 \]

Thus, the final answer is:
\[ \boxed{48} \]