A) \[5-2\sqrt{6}\]
B) \[3\sqrt{3}\]
C) 5
D) \[5+\sqrt{6}\]
Correct Answer: D
Solution :
Since, \[A+B+C=180{}^\circ \] and \[2B=A+C\] \[\Rightarrow \] \[B=60{}^\circ \] \[\therefore \]\[b=9,c=10\]and let\[a=x\] \[\therefore \] \[\cos B=\frac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2ac}\] \[\Rightarrow \] \[\frac{1}{2}=\frac{100+{{x}^{2}}-81}{2.10.x}\] \[\Rightarrow \] \[{{x}^{2}}-10x+19=0\] \[\therefore \] \[x=\frac{10\pm \sqrt{100-76}}{2\times 1}\] \[\Rightarrow \] \[x=5\pm \sqrt{6}\]
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