A) 1: 1: \[\sqrt{2}\]
B) \[1:\sqrt{2}:1\]
C) \[\sqrt{2}:1:1\]
D) None of these
Correct Answer: A
Solution :
From the given relation \[\sin C=\frac{1-\cos A\cos B}{\sin A\sin B}\le 1\] ?..(i) Þ \[1\le \cos A\cos B+\sin A\sin B\] Þ \[\cos (A-B)\ge 1\]; \[\because \cos \theta \not{>}1\] ?..(ii) \ \[A-B=0\] or \[A=B\] Hence from (i), \[\sin C=\frac{1-{{\cos }^{2}}A}{{{\sin }^{2}}A}=\frac{{{\sin }^{2}}A}{{{\sin }^{2}}A}=1\] \ \[C={{90}^{o}}\Rightarrow A+B={{90}^{o}}\]or \[A=B={{45}^{o}}\]{by (ii)} Hence, \[a:b:c\]=\[\sin A:\sin B:\sin C=1:1:\sqrt{2}\].
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