In a `Delta ABC`; inscribed circle with centre `I` touches sides `AB, AC and BC at D, E, F` respectively.Let area of quadrilateral ADIE is 5 units and area of quadrilteral BFID is 10 units. Find the value of `cos(C/2)/sin((A-B)/2)`.

Answer

To solve the problem step by step, we will analyze the given information and use geometric properties of the triangle and its inscribed circle.

### Step 1: Understand the Geometry
We have a triangle \( ABC \) with an inscribed circle that touches the sides \( AB, AC, \) and \( BC \) at points \( D, E, \) and \( F \) respectively. The center of the circle is denoted as \( I \).

### Step 2: Define Areas of Quadrilaterals
We are given:
- Area of quadrilateral \( ADIE = 5 \) units
- Area of quadrilateral \( BFID = 10 \) units

### Step 3: Express Areas in Terms of R and Segments
The area of quadrilateral \( ADIE \) can be expressed as the sum of the areas of triangles \( ADI \) and \( AEI \):
\[ \text{Area of } ADIE = \text{Area of } \triangle ADI + \text{Area of } \triangle AEI \]
Using the formula for the area of a triangle, we have:
\[ \text{Area of } \triangle ADI = \frac{1}{2} R \cdot AD = \frac{1}{2} R \cdot x \]
\[ \text{Area of } \triangle AEI = \frac{1}{2} R \cdot AE = \frac{1}{2} R \cdot x \]
Thus,
\[ \text{Area of } ADIE = \frac{1}{2} R \cdot x + \frac{1}{2} R \cdot x = R \cdot x \]
Setting this equal to 5, we have:
\[ R \cdot x = 5 \quad \text{(1)} \]

### Step 4: Analyze the Second Quadrilateral
For quadrilateral \( BFID \):
\[ \text{Area of } BFID = \text{Area of } \triangle BIE + \text{Area of } \triangle BID \]
Using similar reasoning:
\[ \text{Area of } \triangle BIE = \frac{1}{2} R \cdot BI = \frac{1}{2} R \cdot y \]
\[ \text{Area of } \triangle BID = \frac{1}{2} R \cdot BI = \frac{1}{2} R \cdot y \]
Thus,
\[ \text{Area of } BFID = R \cdot y \]
Setting this equal to 10, we have:
\[ R \cdot y = 10 \quad \text{(2)} \]

### Step 5: Solve for x and y
From equations (1) and (2):
1. \( R \cdot x = 5 \) implies \( x = \frac{5}{R} \)
2. \( R \cdot y = 10 \) implies \( y = \frac{10}{R} \)

### Step 6: Find the Lengths of Sides
The total length \( AB \) can be expressed as:
\[ AB = AD + BD = x + y = \frac{5}{R} + \frac{10}{R} = \frac{15}{R} \]

### Step 7: Apply the Cosine and Sine Formulas
We need to find:
\[ \frac{\cos(\frac{C}{2})}{\sin(\frac{A-B}{2})} \]
Using the formula:
\[ \sin\left(\frac{A-B}{2}\right) = \frac{c}{a-b} \]
Where \( c = AB \), \( a = BC \), and \( b = AC \).

### Step 8: Calculate \( a - b \)
From our earlier definitions:
\[ a = y + z, \quad b = x + z \]
Thus,
\[ a - b = (y + z) - (x + z) = y - x \]
Substituting the values of \( x \) and \( y \):
\[ y - x = \frac{10}{R} - \frac{5}{R} = \frac{5}{R} \]

### Step 9: Final Calculation
Now substituting into the ratio:
\[ \frac{c}{a-b} = \frac{\frac{15}{R}}{\frac{5}{R}} = 3 \]
Thus,
\[ \frac{\cos(\frac{C}{2})}{\sin(\frac{A-B}{2})} = 3 \]

### Final Answer
The value of \( \frac{\cos(\frac{C}{2})}{\sin(\frac{A-B}{2})} \) is \( 3 \).

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