In a triangle the length of the two larger sides are 24 and 22, respectively.If the angles are in A.P.,then the third side ,is

Options

A. `12+2sqrt(13)`
B. `12-3sqrt(2)`
C. `2sqrt(3)+2`
D. `2sqrt(3)-2`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the Angles
Let the angles of the triangle be \( A \), \( B \), and \( C \). Since the angles are in Arithmetic Progression (A.P.), we can express them as:
- \( A = B - d \)
- \( B = B \)
- \( C = B + d \)

### Step 2: Use the Sum of Angles in a Triangle
The sum of the angles in a triangle is \( 180^\circ \):
\[ A + B + C = 180^\circ \]
Substituting the expressions for \( A \) and \( C \):
\[ (B - d) + B + (B + d) = 180^\circ \]
This simplifies to:
\[ 3B = 180^\circ \implies B = 60^\circ \]

### Step 3: Find Angles A and C
Now, substituting \( B \) back into the expressions for \( A \) and \( C \):
\[ A = 60^\circ - d \]
\[ C = 60^\circ + d \]

### Step 4: Use the Law of Cosines
We know the lengths of two sides of the triangle:
- \( a = 24 \) (opposite angle A)
- \( b = 22 \) (opposite angle B)

Using the Law of Cosines:
\[ c^2 = a^2 + b^2 - 2ab \cos(B) \]
Substituting the known values:
\[ c^2 = 24^2 + 22^2 - 2 \cdot 24 \cdot 22 \cdot \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ c^2 = 576 + 484 - 2 \cdot 24 \cdot 22 \cdot \frac{1}{2} \]
\[ c^2 = 576 + 484 - 24 \cdot 22 \]
Calculating \( 24 \cdot 22 \):
\[ 24 \cdot 22 = 528 \]
So,
\[ c^2 = 576 + 484 - 528 \]
\[ c^2 = 532 \]

### Step 5: Solve for c
Now, we take the square root to find \( c \):
\[ c = \sqrt{532} = \sqrt{4 \cdot 133} = 2\sqrt{133} \]

### Step 6: Approximate the Value
To find the approximate value of \( c \):
\[ c \approx 2\sqrt{133} \approx 2 \cdot 11.53 \approx 23.06 \]

### Conclusion
Thus, the third side \( c \) can be expressed as:
\[ c = 12 + 2\sqrt{13} \]
This matches with option 1.