If in `DeltaABC, a = 5, b = 4 and cos (A - B) = 31/32`, then side c is

Options

A. `sqrt(6)`
B. `6sqrt(6)`
C. 6
D. `(216)^(1//4)`

Answer

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

### Step 1: Use the cosine of the angle difference
We know that:
\[ \cos(A - B) = \cos A \cos B + \sin A \sin B \]
Since we have \( \cos(A - B) = \frac{31}{32} \), we will need to find \( \sin A \) and \( \sin B \) using the sides \( a \) and \( b \).

### Step 2: Use the tangent half-angle formula
Using the formula for \( \tan\left(\frac{A - B}{2}\):
\[ \tan\left(\frac{A - B}{2}\right) = \sqrt{\frac{1 - \cos(A - B)}{1 + \cos(A - B)}} \]
Substituting \( \cos(A - B) = \frac{31}{32} \):
\[ \tan\left(\frac{A - B}{2}\right) = \sqrt{\frac{1 - \frac{31}{32}}{1 + \frac{31}{32}} = \sqrt{\frac{\frac{1}{32}}{\frac{63}{32}} = \sqrt{\frac{1}{63}} = \frac{1}{\sqrt{63}} \]

### Step 3: Relate \( \tan\left(\frac{A - B}{2}\) to \( A \) and \( B \)
From the tangent half-angle formula:
\[ \tan\left(\frac{A - B}{2}\right) = \frac{a - b}{a + b} \cdot \cot\left(\frac{C}{2} \]
Substituting \( a = 5 \) and \( b = 4 \):
\[ \frac{5 - 4}{5 + 4} \cdot \cot\left(\frac{C}{2} = \frac{1}{9} \cdot \cot\left(\frac{C}{2} \]
Setting this equal to \( \frac{1}{\sqrt{63}} \):
\[ \frac{1}{9} \cdot \cot\left(\frac{C}{2} = \frac{1}{\sqrt{63}} \]
Solving for \( \cot\left(\frac{C}{2} \):
\[ \cot\left(\frac{C}{2} = \frac{9}{\sqrt{63}} = \frac{9\sqrt{63}}{63} = \frac{3\sqrt{7}}{7} \]

### Step 4: Find \( \tan\left(\frac{C}{2}\)
Using the identity \( \tan\left(\frac{C}{2}\) is the reciprocal of \( \cot\left(\frac{C}{2}\):
\[ \tan\left(\frac{C}{2} = \frac{7}{3\sqrt{7}} = \frac{7\sqrt{7}}{21} = \frac{\sqrt{7}}{3} \]

### Step 5: Use the cosine rule to find \( c \)
Using the cosine rule:
\[ c^2 = a^2 + b^2 - 2ab \cos C \]
We need to find \( \cos C \) using:
\[ \cos C = \frac{1 - \tan^2\left(\frac{C}{2}}{1 + \tan^2\left(\frac{C}{2}} \]
Substituting \( \tan\left(\frac{C}{2} = \frac{\sqrt{7}}{3}: \[ \cos C = \frac{1 - \left(\frac{7}{9}\right)}{1 + \left(\frac{7}{9}\right)} = \frac{2}{16/9} = \frac{1}{8} \] ### Step 6: Substitute values into the cosine rule Now substituting \( a = 5 \), \( b = 4 \), and \( \cos C = \frac{1}{8} \):
\[ c^2 = 5^2 + 4^2 - 2 \cdot 5 \cdot 4 \cdot \frac{1}{8} \]
Calculating:
\[ c^2 = 25 + 16 - \frac{40}{8} = 25 + 16 - 5 = 36 \]
Thus,
\[ c = \sqrt{36} = 6 \]

### Final Answer
The length of side \( c \) is \( 6 \).

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