In a triangle ABC,with usual notations,if a=5 | vedclass

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(A) Using the Napier's Analogy,we have $	an\left(\frac{A-B}{2}ight) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}ight)$.Given $a=5, b=4$,so $\frac{a-b}

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$6$

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(A) Using the Napier's Analogy,we have $	an\left(\frac{A-B}{2}ight) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}ight)$.Given $a=5, b=4$,so $\frac{a-b}{a+b} = \frac{5-4}{5+4} = \frac{1}{9}$.Also,$\cos(A-B) = 2\cos^2\left(\frac{A-B}{2}ight) - 1 = \frac{31}{32}$,which gives $\cos^2\left(\frac{A-B}{2}ight) = \frac{63}{64}$.Thus,$	an^2\left(\frac{A-B}{2}ight) = \sec^2\left(\frac{A-B}{2}ight) - 1 = \frac{64}{63} - 1 = \frac{1}{63}$.So,$	an\left(\frac{A-B}{2}ight) = \frac{1}{\sqrt{63}} = \frac{1}{3\sqrt{7}}$.Substituting this into the Napier's formula: $\frac{1}{3\sqrt{7}} = \frac{1}{9} \cot\left(\frac{C}{2}ight)$,which implies $\cot\left(\frac{C}{2}ight) = \frac{9}{3\sqrt{7}} = \frac{3}{\sqrt{7}}$.Then $	an^2\left(\frac{C}{2}ight) = \frac{7}{9}$,so $\cos C = \frac{1-	an^2(C/2)}{1+	an^2(C/2)} = \frac{1-7/9}{1+7/9} = \frac{2/9}{16/9} = \frac{1}{8}$.Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C = 5^2 + 4^2 - 2(5)(4)(\frac{1}{8}) = 25 + 16 - 5 = 36$.Therefore,$c = 6$.

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

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