A) \[4\]
B) \[5\]
C) \[6-\sqrt{6}\]
D) \[5+\sqrt{6}\]
Correct Answer: D
Solution :
Given longer sides are \[c=10\] and \[b=9.\] Using sine rule, \[\frac{\sin \,B}{b}=\frac{\sin \,C}{c}\] But angles in AP, \[\therefore \] \[2B=A+C\] Also, \[A+B+C={{180}^{o}}\] \[3B={{180}^{o}}\Rightarrow B={{60}^{o}}\] \[\therefore \] \[\frac{\sin \,{{60}^{o}}}{9}=\frac{\sin \,C}{10}\] \[\Rightarrow \] \[\frac{\sqrt{3}}{2\times 9}=\frac{\sin \,C}{10}\] \[\because \] \[\cos \,B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] \[\therefore \] \[\cos \,{{60}^{o}}=\frac{{{a}^{2}}+{{(10)}^{2}}-{{(9)}^{2}}}{2a\times 10}\] \[\Rightarrow \] \[\frac{1}{2}=\frac{{{a}^{2}}+19}{20a}\] \[\Rightarrow \] \[10\,a={{a}^{2}}+19\] \[{{a}^{2}}-10a+19=0\] \[\therefore \] \[a=\frac{10\pm \,\sqrt{{{(10)}^{2}}-4\times 19}}{2(1)}\] \[=\frac{10\pm \,\sqrt{24}}{2}\] \[=\frac{10\pm \,2\sqrt{6}}{2}\] \[=5\pm \sqrt{6}\]
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