In a triangle , the lengths of two larger are 10 cm and 9 cm . If the angles of the triangle are in AP, then the length of the third side is

Options

A. ` sqrt(5) - sqrt(6)`
B. `sqrt(5)+sqrt(6)`
C. `sqrt(5)pmsqrt(6)`
D. `5 pm sqrt(6)`

Explanation

To solve the problem, we need to find the length of the third side of a triangle where two sides are given as 10 cm and 9 cm, and the angles are in arithmetic progression (AP).

### Step-by-Step Solution:

1. **Identify the sides and angles of the triangle**:
Let the sides of the triangle be:
- \( a = 10 \) cm (opposite angle A)
- \( b = 9 \) cm (opposite angle B)
- \( c = x \) cm (opposite angle C)

Since the angles are in AP, we can denote them as:
- Angle A = \( \theta + d \)
- Angle B = \( \theta \)
- Angle C = \( \theta - d \)

2. **Use the property of angles in a triangle**:
The sum of the angles in a triangle is \( 180^\circ \):
\[ (\theta + d) + \theta + (\theta - d) = 180^\circ \]
Simplifying this gives:
\[ 3\theta = 180^\circ \implies \theta = 60^\circ \]

3. **Substituting the value of \( \theta \)**:
Now we can substitute \( \theta \) back into the angles:
- Angle A = \( 60^\circ + d \)
- Angle B = \( 60^\circ \)
- Angle C = \( 60^\circ - d \)

4. **Using the Law of Cosines**:
We can apply the Law of Cosines to find the third side \( c \):
\[ c^2 = a^2 + b^2 - 2ab \cdot \cos(B) \]
Substituting the known values:
\[ x^2 = 10^2 + 9^2 - 2 \cdot 10 \cdot 9 \cdot \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ x^2 = 100 + 81 - 2 \cdot 10 \cdot 9 \cdot \frac{1}{2} \]
Simplifying this:
\[ x^2 = 100 + 81 - 90 = 91 \]

5. **Finding the value of \( x \)**:
Taking the square root of both sides:
\[ x = \sqrt{91} \]

6. **Final answer**:
The length of the third side is:
\[ x \approx 9.54 \text{ cm} \]