A) 6
B) 7
C) 9
D) None of these
Correct Answer: A
Solution :
We have \[\tan \left( \frac{A-B}{2} \right)=\sqrt{\frac{1-\cos (A-B)}{1+\cos (A-B)}}\] \[=\sqrt{\frac{1-(31/32)}{1+(31/32)}}=\frac{1}{\sqrt{63}}\] Þ \[\frac{a-b}{a+b}\cot \frac{C}{2}=\frac{1}{\sqrt{63}}\] Þ \[\frac{1}{9}\cot \frac{C}{2}=\frac{1}{\sqrt{63}}\Rightarrow \tan \frac{C}{2}=\frac{\sqrt{7}}{3}\] Now \[\cos C=\frac{1-{{\tan }^{2}}(C/2)}{1+{{\tan }^{2}}(C/2)}\]Þ \[\cos C=\frac{1-(7/9)}{1+(7/9)}=\frac{1}{8}\] \\[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C\] Þ \[{{c}^{2}}=25+16-40\times \frac{1}{8}=36\Rightarrow c=6\].
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