Question

In a triangle ABC , with usual notations, if \(\mathrm{a}=5, \mathrm{~b}=4, \cos (\mathrm{~A}-\mathrm{B})=\frac{31}{32}\), then \(\mathrm{c}=\)

• A. 6
• B. 7
• C. 5
• D. 2

Hint:

Convert cos(A - B) into tan((A - B)/2), apply Napier's analogy to get C, then use the cosine rule.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Use \(\tan^2\dfrac{A-B}{2} = \dfrac{1 - \cos(A-B)}{1 + \cos(A-B)}\).

Step 2: Compute the half-angle tangent
\(\tan^2\dfrac{A-B}{2} = \dfrac{1/32}{63/32} = \dfrac{1}{63}\), so \(\tan\dfrac{A-B}{2} = \dfrac{1}{3\sqrt{7}}\).

Step 3: Napier's analogy
\(\tan\dfrac{A-B}{2} = \dfrac{a-b}{a+b}\cot\dfrac{C}{2} = \dfrac{1}{9}\cot\dfrac{C}{2}\), so \(\cot\dfrac{C}{2} = \dfrac{3}{\sqrt{7}}\).

Step 4: Find cos C
\(\tan^2\dfrac{C}{2} = \dfrac{7}{9}\), so \(\cos C = \dfrac{1 - 7/9}{1 + 7/9} = \dfrac{1}{8}\).

Step 5: Cosine rule
\(c^2 = 25 + 16 - 2\cdot5\cdot4\cdot\dfrac{1}{8} = 41 - 5 = 36\), so \(c = 6\).

Final Answer:
The third side is 6. \[ \boxed{\text{(A) }6} \]