In a triangle ABC , if `r_(1)=3ands=4` , then cosA =

Options

A. `(4)/(5)`
B. `(3)/(5)`
C. `(24)/(25)`
D. `(7)/(25)`

Answer

To solve for \( \cos A \) in triangle \( ABC \) given \( r_1 = 3 \) and \( s = 4 \), we can follow these steps:

### Step 1: Understand the Relationship
The relationship between the inradius \( r_1 \), the area \( \Delta \), and the semi-perimeter \( s \) is given by the formula:
\[ r_1 = \frac{\Delta}{s} \]
However, we also have another formula involving the tangent of half the angle:
\[ r_1 = \frac{\Delta}{s - a} = \frac{s \cdot \tan \frac{A}{2}}{2} \]
For our case, we will use:
\[ r_1 = s \cdot \tan \frac{A}{2} \]

### Step 2: Substitute Known Values
Substituting the known values \( r_1 = 3 \) and \( s = 4 \):
\[ 3 = 4 \cdot \tan \frac{A}{2} \]

### Step 3: Solve for \( \tan \frac{A}{2} \)
Rearranging the equation gives:
\[ \tan \frac{A}{2} = \frac{3}{4} \]

### Step 4: Use the Half-Angle Formula for Cosine
We can use the half-angle formula for cosine:
\[ \cos A = \frac{1 - \tan^2 \frac{A}{2}}{1 + \tan^2 \frac{A}{2}} \]
Substituting \( \tan \frac{A}{2} = \frac{3}{4} \):
\[ \cos A = \frac{1 - \left(\frac{3}{4}\right)^2}{1 + \left(\frac{3}{4}\right)^2} \]

### Step 5: Calculate \( \tan^2 \frac{A}{2} \)
Calculating \( \tan^2 \frac{A}{2} \):
\[ \tan^2 \frac{A}{2} = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \]

### Step 6: Substitute Back into the Cosine Formula
Substituting \( \tan^2 \frac{A}{2} \) into the cosine formula:
\[ \cos A = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \frac{\frac{16 - 9}{16}}{\frac{16 + 9}{16}} = \frac{7/16}{25/16} \]

### Step 7: Simplify the Expression
This simplifies to:
\[ \cos A = \frac{7}{25} \]

### Final Answer
Thus, the value of \( \cos A \) is:
\[ \cos A = \frac{7}{25} \]