In triangle ABC,if a=5 and tan frac{A-B}{2}=f | vedclass

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(C) Using the Napier's analogy,we have $	an \frac{A-B}{2} = \frac{a-b}{a+b} \cot \frac{C}{2}$ and $	an \frac{A+B}{2} = \cot \frac{C}{2}$.Given $	an

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$4$

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(C) Using the Napier's analogy,we have $	an \frac{A-B}{2} = \frac{a-b}{a+b} \cot \frac{C}{2}$ and $	an \frac{A+B}{2} = \cot \frac{C}{2}$.Given $	an \frac{A-B}{2} = \frac{1}{4} 	an \frac{A+B}{2}$,we substitute these values:$\frac{a-b}{a+b} \cot \frac{C}{2} = \frac{1}{4} \cot \frac{C}{2}$$\Rightarrow \frac{a-b}{a+b} = \frac{1}{4}$$\Rightarrow 4(a-b) = a+b$$\Rightarrow 4a - 4b = a + b$$\Rightarrow 3a = 5b$Given $a=5$,we have $3(5) = 5b \Rightarrow b=3$.Now,$\sqrt{a^2-b^2} = \sqrt{5^2-3^2} = \sqrt{25-9} = \sqrt{16} = 4$.

In $\triangle ABC$,if $a=5$ and $\tan \frac{A-B}{2}=\frac{1}{4} \tan \frac{A+B}{2}$,then $\sqrt{a^2-b^2}=$

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