In a triangle ABC,if a=7, c=11, cos A=frac{17 | vedclass

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(C) Given $a=7, c=11, \cos A=\frac{17}{22}, \cos C=\frac{1}{14}$.Using the law of tangents:$	an \left(\frac{C-A}{2}ight) = \left(\frac{c-a}{c+a}i

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$2$

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(C) Given $a=7, c=11, \cos A=\frac{17}{22}, \cos C=\frac{1}{14}$.Using the law of tangents:$	an \left(\frac{C-A}{2}ight) = \left(\frac{c-a}{c+a}ight) \cot \left(\frac{B}{2}ight)$.Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$:$\frac{17}{22} = \frac{b^2+121-49}{2b(11)} \implies \frac{17}{22} = \frac{b^2+72}{22b} \implies 17b = b^2+72 \implies b^2-17b+72=0$.Solving the quadratic equation: $(b-8)(b-9)=0$,so $b=8$ or $b=9$.For $b=9$,the expression becomes:$b 	an \frac{B}{2} 	an \frac{C-A}{2} = b 	an \frac{B}{2} \left(\frac{c-a}{c+a}ight) \cot \frac{B}{2} = b \left(\frac{c-a}{c+a}ight) = 9 \left(\frac{11-7}{11+7}ight) = 9 \left(\frac{4}{18}ight) = 2$.Thus,the correct option is $C$.

In a triangle $ABC$,if $a=7, c=11, \cos A=\frac{17}{22}, \cos C=\frac{1}{14}$,then $b \tan \frac{B}{2} \tan \frac{C-A}{2} =$

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