A) \[1:4:3\]
B) \[4:1:3\]
C) \[4:3:1\]
D) \[3:4:1\]
Correct Answer: B
Solution :
\[\cos 120{}^\circ =\frac{{{x}^{2}}+{{x}^{2}}-A{{B}^{2}}}{2{{x}^{2}}}\] Þ \[\frac{2{{x}^{2}}-A{{B}^{2}}}{2{{x}^{2}}}=\frac{-1}{2}\] Þ \[4{{x}^{2}}-2A{{B}^{2}}=-2{{x}^{2}}\] Þ \[3{{x}^{2}}=A{{B}^{2}}\]Þ\[AB=x\sqrt{3}\] Þ \[{{a}^{2}}:{{b}^{2}}:{{c}^{2}}={{(2x)}^{2}}:{{x}^{2}}:{{(x\sqrt{3})}^{2}}\] = \[4{{x}^{2}}:{{x}^{2}}:3{{x}^{2}}\] = \[4:1:3\].You need to login to perform this action.
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