In a triangle ABC if `a =5,b= 4,cos (A - B) = 31/32`, then the third side c =

Options

A. 4
B. 6
C. 8
D. 10

Answer

To find the third side \( c \) of triangle \( ABC \) given \( a = 5 \), \( b = 4 \), and \( \cos(A - B) = \frac{31}{32} \), we can follow these steps:

### Step 1: Use the formula for \( \cos(A - B) \)
The formula for \( \cos(A - B) \) is:

\[ \cos(A - B) = \cos A \cos B + \sin A \sin B \]

However, we will use a different approach involving the tangent half-angle formula. We know that:

\[ \cos(A - B) = \frac{1 - \tan^2\left(\frac{A - B}{2}\right)}{1 + \tan^2\left(\frac{A - B}{2}\right)} \]

Let \( y = \tan\left(\frac{A - B}{2}\right) \). Then we can rewrite the equation as:

\[ \frac{1 - y^2}{1 + y^2} = \frac{31}{32} \]

### Step 2: Cross-multiply and solve for \( y \)
Cross-multiplying gives us:

\[ 32(1 - y^2) = 31(1 + y^2) \]

Expanding both sides:

\[ 32 - 32y^2 = 31 + 31y^2 \]

Rearranging terms:

\[ 32 - 31 = 32y^2 + 31y^2 \]

\[ 1 = 63y^2 \]

Thus,

\[ y^2 = \frac{1}{63} \]

### Step 3: Find \( \tan\left(\frac{A - B}{2}\right) \)
Taking the square root gives:

\[ y = \tan\left(\frac{A - B}{2}\right) = \frac{1}{\sqrt{63}} \]

### Step 4: Use the tangent half-angle formula
Using the formula:

\[ \tan\left(\frac{A - B}{2}\right) = \frac{a - b}{a + b} \cdot \cot\left(\frac{C}{2}\right) \]

Substituting the known values \( a = 5 \) and \( b = 4 \):

\[ \frac{1}{\sqrt{63}} = \frac{5 - 4}{5 + 4} \cdot \cot\left(\frac{C}{2}\right) \]

This simplifies to:

\[ \frac{1}{\sqrt{63}} = \frac{1}{9} \cdot \cot\left(\frac{C}{2}\right) \]

### Step 5: Solve for \( \cot\left(\frac{C}{2}\right) \)
Cross-multiplying gives:

\[ \cot\left(\frac{C}{2}\right) = 9 \cdot \frac{1}{\sqrt{63}} = \frac{9}{\sqrt{63}} = \frac{9\sqrt{63}}{63} = \frac{3\sqrt{7}}{21} = \frac{3\sqrt{7}}{21} \]

### Step 6: Find \( \cos C \)
Using the formula for \( \cos C \):

\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

We can express \( \cos C \) in terms of \( \tan\left(\frac{C}{2}\right) \):

\[ \cos C = \frac{1 - \tan^2\left(\frac{C}{2}\right)}{1 + \tan^2\left(\frac{C}{2}\right)} \]

### Step 7: Substitute \( \tan\left(\frac{C}{2}\right) \)
From the cotangent value, we can find \( \tan\left(\frac{C}{2}\right) \):

\[ \tan\left(\frac{C}{2}\right) = \frac{1}{\cot\left(\frac{C}{2}\right)} = \frac{\sqrt{63}}{9} \]

Now substituting \( \tan\left(\frac{C}{2}\right) \) into the cosine formula:

\[ \cos C = \frac{1 - \left(\frac{\sqrt{63}}{9}\right)^2}{1 + \left(\frac{\sqrt{63}}{9}\right)^2} \]

Calculating this gives:

\[ \cos C = \frac{1 - \frac{63}{81}}{1 + \frac{63}{81}} = \frac{\frac{18}{81}}{\frac{144}{81}} = \frac{18}{144} = \frac{1}{8} \]

### Step 8: Set up the equation for \( c \)
Now substituting back into the cosine formula:

\[ \frac{1}{8} = \frac{5^2 + 4^2 - c^2}{2 \cdot 5 \cdot 4} \]

This simplifies to:

\[ \frac{1}{8} = \frac{25 + 16 - c^2}{40} \]

Cross-multiplying gives:

\[ 40 = 8(41 - c^2) \]

Expanding and simplifying:

\[ 40 = 328 - 8c^2 \]

Rearranging gives:

\[ 8c^2 = 328 - 40 \]

\[ 8c^2 = 288 \]

\[ c^2 = 36 \]

Taking the square root:

\[ c = 6 \]

### Final Answer
Thus, the third side \( c \) is:

\[ \boxed{6} \]