In `Delta ABC`, angle A is `120^(@), BC + CA = 20, and AB + BC = 21` Find the length of the side BC

Explanation

To solve the problem step-by-step, we will denote the sides of triangle ABC as follows:
- Let \( AB = c \)
- Let \( BC = a \)
- Let \( CA = b \)

Given:
1. \( \angle A = 120^\circ \)
2. \( BC + CA = 20 \) → \( a + b = 20 \)
3. \( AB + BC = 21 \) → \( c + a = 21 \)

We need to find the length of side \( BC \) (which is \( a \)).

### Step 1: Express \( b \) and \( c \) in terms of \( a \)
From the equations:
- From \( a + b = 20 \), we can express \( b \) as:
\[ b = 20 - a \]
- From \( c + a = 21 \), we can express \( c \) as:
\[ c = 21 - a \]

### Step 2: Use the Cosine Rule
Using the cosine rule for triangle ABC:
\[ c^2 = a^2 + b^2 - 2ab \cos(A) \]
Substituting \( A = 120^\circ \) (where \( \cos(120^\circ) = -\frac{1}{2} \)):
\[ c^2 = a^2 + b^2 + ab \]

### Step 3: Substitute \( b \) and \( c \)
Substituting \( b = 20 - a \) and \( c = 21 - a \) into the cosine rule equation:
\[ (21 - a)^2 = a^2 + (20 - a)^2 + a(20 - a) \]

### Step 4: Expand and simplify
Expanding both sides:
- Left side:
\[ (21 - a)^2 = 441 - 42a + a^2 \]

- Right side:
\[ a^2 + (20 - a)^2 + a(20 - a) = a^2 + (400 - 40a + a^2) + (20a - a^2) \]
\[ = a^2 + 400 - 40a + a^2 + 20a - a^2 = a^2 - 20a + 400 \]

### Step 5: Set the equation
Setting the left side equal to the right side:
\[ 441 - 42a + a^2 = a^2 - 20a + 400 \]

### Step 6: Cancel \( a^2 \) and rearrange
Cancelling \( a^2 \) from both sides:
\[ 441 - 42a = -20a + 400 \]
Rearranging gives:
\[ 441 - 400 = 42a - 20a \]
\[ 41 = 22a \]
\[ a = \frac{41}{22} \]

### Step 7: Calculate \( a \)
Calculating the value:
\[ a = 1.8636 \text{ (approximately)} \]

### Conclusion
The length of side \( BC \) is approximately \( 1.86 \).