In a triangle ABC,l(AB)=sqrt{23} units,l(BC)= | vedclass

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(B) Let $a, b, c$ be the side lengths opposite to vertices $A, B, C$ respectively. Here,$a = 3$,$b = 4$,and $c = \sqrt{23}$.Using the cotangent rule,$

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$2$

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(B) Let $a, b, c$ be the side lengths opposite to vertices $A, B, C$ respectively. Here,$a = 3$,$b = 4$,and $c = \sqrt{23}$.Using the cotangent rule,$\cot A = \frac{b^2+c^2-a^2}{4\Delta}$,$\cot B = \frac{a^2+c^2-b^2}{4\Delta}$,and $\cot C = \frac{a^2+b^2-c^2}{4\Delta}$,where $\Delta$ is the area of the triangle.Then,$\frac{\cot A+\cot C}{\cot B} = \frac{\frac{b^2+c^2-a^2}{4\Delta} + \frac{a^2+b^2-c^2}{4\Delta}}{\frac{a^2+c^2-b^2}{4\Delta}} = \frac{b^2+c^2-a^2+a^2+b^2-c^2}{a^2+c^2-b^2} = \frac{2b^2}{a^2+c^2-b^2}$.Substituting the values: $a^2 = 9$,$b^2 = 16$,$c^2 = 23$.$\frac{2(16)}{9+23-16} = \frac{32}{16} = 2$.

In a triangle $ABC$,$l(AB)=\sqrt{23}$ units,$l(BC)=3$ units,$l(CA)=4$ units,then $\frac{\cot A+\cot C}{\cot B}$ is

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