In a Delta ABC,a = 5, b = 4 and cos(A - B) = | vedclass

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(A) Using the Napier's analogy: $	an \left( \frac{A - B}{2} ight) = \sqrt{\frac{1 - \cos(A - B)}{1 + \cos(A - B)}} = \sqrt{\frac{1 - 31/32}{1 + 31/

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$6$

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(A) Using the Napier's analogy: $	an \left( \frac{A - B}{2} ight) = \sqrt{\frac{1 - \cos(A - B)}{1 + \cos(A - B)}} = \sqrt{\frac{1 - 31/32}{1 + 31/32}} = \sqrt{\frac{1/32}{63/32}} = \frac{1}{\sqrt{63}} = \frac{1}{3\sqrt{7}}$.By Napier's formula,$\frac{a - b}{a + b} \cot \frac{C}{2} = 	an \left( \frac{A - B}{2} ight)$.Substituting the values: $\frac{5 - 4}{5 + 4} \cot \frac{C}{2} = \frac{1}{3\sqrt{7}}$ $\Rightarrow \frac{1}{9} \cot \frac{C}{2} = \frac{1}{3\sqrt{7}}$ $\Rightarrow \cot \frac{C}{2} = \frac{3}{\sqrt{7}}$.Thus,$	an \frac{C}{2} = \frac{\sqrt{7}}{3}$.Using $\cos C = \frac{1 - 	an^2(C/2)}{1 + 	an^2(C/2)} = \frac{1 - 7/9}{1 + 7/9} = \frac{2/9}{16/9} = \frac{1}{8}$.Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C = 25 + 16 - 2(5)(4) \left( \frac{1}{8} ight) = 41 - 40 \left( \frac{1}{8} ight) = 41 - 5 = 36$.Therefore,$c = \sqrt{36} = 6$.

In a $\Delta ABC$,$a = 5, b = 4$ and $\cos(A - B) = \frac{31}{32}$,then side $c$ is equal to

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