In a triangle ABC,let AB = sqrt{23},BC = 3,an | vedclass

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(B) Given sides are $c = AB = \sqrt{23}$,$a = BC = 3$,and $b = CA = 4$.Using the formula for $\cot$ in terms of sides and area $\Delta$:$\cot A = \fra

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$2$

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(B) Given sides are $c = AB = \sqrt{23}$,$a = BC = 3$,and $b = CA = 4$.Using the formula for $\cot$ in terms of sides and area $\Delta$:$\cot A = \frac{b^2 + c^2 - a^2}{4\Delta}$,$\cot B = \frac{a^2 + c^2 - b^2}{4\Delta}$,and $\cot C = \frac{a^2 + b^2 - c^2}{4\Delta}$.Now,calculate the expression:$\frac{\cot A + \cot C}{\cot B} = \frac{\frac{b^2 + c^2 - a^2}{4\Delta} + \frac{a^2 + b^2 - c^2}{4\Delta}}{\frac{a^2 + c^2 - b^2}{4\Delta}}$$= \frac{b^2 + c^2 - a^2 + a^2 + b^2 - c^2}{a^2 + c^2 - b^2}$$= \frac{2b^2}{a^2 + c^2 - b^2}$Substitute the values $a = 3$,$b = 4$,$c = \sqrt{23}$:$= \frac{2(4^2)}{3^2 + (\sqrt{23})^2 - 4^2}$$= \frac{2(16)}{9 + 23 - 16}$$= \frac{32}{32 - 16} = \frac{32}{16} = 2$.

In a triangle $ABC$,let $AB = \sqrt{23}$,$BC = 3$,and $CA = 4$. Then the value of $\frac{\cot A + \cot C}{\cot B}$ is:

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