In a triangle ABC,if c=9, s=10 and Delta=10sq | vedclass

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(A) Given for triangle $ABC$: $c=9, s=10, \Delta=10\sqrt{2}$.Using Napier's Analogy: $	an\left(\frac{A-B}{2}ight) = \frac{a-b}{a+b}\cot\left(\frac{

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$a\left[1-\sqrt{2}	an\left(\frac{A-B}{2}ight)ight]$

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(A) Given for triangle $ABC$: $c=9, s=10, \Delta=10\sqrt{2}$.Using Napier's Analogy: $	an\left(\frac{A-B}{2}ight) = \frac{a-b}{a+b}\cot\left(\frac{C}{2}ight)$.We know $\cot\left(\frac{C}{2}ight) = \frac{s(s-c)}{\Delta} = \frac{10(10-9)}{10\sqrt{2}} = \frac{1}{\sqrt{2}}$.Thus,$	an\left(\frac{A-B}{2}ight) = \frac{a-b}{a+b} \cdot \frac{1}{\sqrt{2}}$,which implies $\sqrt{2}	an\left(\frac{A-B}{2}ight) = \frac{a-b}{a+b}$.Now,substitute this into the expression $b\left[1+\sqrt{2}	an\left(\frac{A-B}{2}ight)ight]$:$= b\left[1 + \frac{a-b}{a+b}ight] = b\left[\frac{a+b+a-b}{a+b}ight] = b\left[\frac{2a}{a+b}ight]$.Rearranging the terms: $a\left[\frac{2b}{a+b}ight] = a\left[\frac{(a+b)-(a-b)}{a+b}ight] = a\left[1 - \frac{a-b}{a+b}ight]$.Substituting back $\frac{a-b}{a+b} = \sqrt{2}	an\left(\frac{A-B}{2}ight)$,we get:$= a\left[1 - \sqrt{2}	an\left(\frac{A-B}{2}ight)ight]$.

In a triangle $ABC$,if $c=9, s=10$ and $\Delta=10\sqrt{2}$,then $b\left[1+\sqrt{2}\tan\left(\frac{A-B}{2}\right)\right]=$

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