In triangle `ABC, a=5, b=4` and `cos(A+B)=(31)/(32)` In this triangle,`c=`

Options

A. 3
B. 6
C. 7
D. 9

Answer

To solve the problem step by step, we will follow the reasoning and calculations outlined in the video transcript.

### Step 1: Understand the Given Information
We are given:
- Side \( a = 5 \)
- Side \( b = 4 \)
- \( \cos(A + B) = \frac{31}{32} \)

We need to find the length of side \( c \).

### Step 2: Use the Relationship Between Angles
In triangle \( ABC \), we know that:
\[ A + B + C = 180^\circ \]
This implies:
\[ A + B = 180^\circ - C \]
Taking the cosine of both sides, we have:
\[ \cos(A + B) = \cos(180^\circ - C) = -\cos(C) \]

### Step 3: Substitute the Given Cosine Value
We can substitute the given value of \( \cos(A + B) \):
\[ \frac{31}{32} = -\cos(C) \]
This leads to:
\[ \cos(C) = -\frac{31}{32} \]

### Step 4: Use the Cosine Rule
The cosine rule states:
\[ \cos(C) = \frac{a^2 + b^2 - c^2}{2ab} \]
Substituting the known values:
\[ -\frac{31}{32} = \frac{5^2 + 4^2 - c^2}{2 \cdot 5 \cdot 4} \]
Calculating \( 5^2 + 4^2 \):
\[ 5^2 = 25, \quad 4^2 = 16 \quad \Rightarrow \quad 25 + 16 = 41 \]
Thus, we have:
\[ -\frac{31}{32} = \frac{41 - c^2}{40} \]

### Step 5: Cross-Multiply to Solve for \( c^2 \)
Cross-multiplying gives:
\[ -31 \cdot 40 = 32(41 - c^2) \]
Calculating the left side:
\[ -1240 = 32 \cdot 41 - 32c^2 \]
Calculating \( 32 \cdot 41 \):
\[ 32 \cdot 41 = 1312 \]
So we have:
\[ -1240 = 1312 - 32c^2 \]

### Step 6: Rearranging the Equation
Rearranging the equation to isolate \( c^2 \):
\[ 32c^2 = 1312 + 1240 \]
Calculating the right side:
\[ 32c^2 = 2552 \]
Dividing by 32:
\[ c^2 = \frac{2552}{32} = 79 \]

### Step 7: Finding \( c \)
Taking the square root:
\[ c = \sqrt{79} \approx 8.89 \]

### Conclusion
Thus, the length of side \( c \) is approximately \( 8.89 \).