In `DeltaABC, tanA=2, tanB=(3)/(2) and c=sqrt(65)`, then circumradius of the triangle is : (a) 65 (b) `(65)/(7)` (c) `(65)/(14)` (d) none of these

Options

A. 65
B. `(65)/(7)`
C. `(65)/(14)`
D. none of these

Answer

To find the circumradius \( R \) of triangle \( ABC \) given \( \tan A = 2 \), \( \tan B = \frac{3}{2} \), and \( c = \sqrt{65} \), we can follow these steps:

### Step 1: Find \( \tan C \)
Using the formula for the tangent of the sum of two angles:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Substituting the values:
\[ \tan A = 2, \quad \tan B = \frac{3}{2} \]
\[ \tan(A + B) = \frac{2 + \frac{3}{2}}{1 - 2 \cdot \frac{3}{2}} = \frac{\frac{4}{2} + \frac{3}{2}}{1 - 3} = \frac{\frac{7}{2}}{-2} = -\frac{7}{4} \]
Thus, we have:
\[ \tan C = -\tan(A + B) = \frac{7}{4} \]

### Step 2: Find \( \cot C \)
Using the relationship between tangent and cotangent:
\[ \cot C = \frac{1}{\tan C} = \frac{4}{7} \]

### Step 3: Find \( \csc^2 C \)
Using the identity:
\[ 1 + \cot^2 C = \csc^2 C \]
Substituting \( \cot C \):
\[ 1 + \left(\frac{4}{7}\right)^2 = 1 + \frac{16}{49} = \frac{49 + 16}{49} = \frac{65}{49} \]
Thus,
\[ \csc^2 C = \frac{65}{49} \]

### Step 4: Find \( \csc C \)
Taking the square root:
\[ \csc C = \sqrt{\frac{65}{49}} = \frac{\sqrt{65}}{7} \]

### Step 5: Find \( \sin C \)
Using the relationship:
\[ \sin C = \frac{1}{\csc C} = \frac{7}{\sqrt{65}} \]

### Step 6: Use the Extended Sine Rule
The extended sine rule states:
\[ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = \frac{1}{2R} \]
We know \( c = \sqrt{65} \) and \( \sin C = \frac{7}{\sqrt{65}} \):
\[ \frac{\sin C}{c} = \frac{\frac{7}{\sqrt{65}}}{\sqrt{65}} = \frac{7}{65} \]

### Step 7: Set up the equation for \( R \)
From the extended sine rule:
\[ \frac{7}{65} = \frac{1}{2R} \]
Cross-multiplying gives:
\[ 7 \cdot 2R = 65 \implies 14R = 65 \implies R = \frac{65}{14} \]

### Conclusion
Thus, the circumradius \( R \) of triangle \( ABC \) is:
\[ \boxed{\frac{65}{14}} \]