A) \[{{90}^{o}},{{30}^{o}}\]
B) \[{{60}^{o}},{{60}^{o}}\]
C) \[{{30}^{o}},{{90}^{o}}\]
D) \[{{60}^{o}},{{45}^{o}}\]
Correct Answer: C
Solution :
Given, \[C={{60}^{o}},a=2,b=4\] Þ \[\cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\]or \[ab={{a}^{2}}+{{b}^{2}}-{{c}^{2}}\] Þ \[8=4+16-{{c}^{2}}\]Þ \[{{c}^{2}}=12\Rightarrow c=\sqrt{12}=2\sqrt{3}\]. Þ \[\sin A=\frac{a\sin C}{c}=\frac{2\text{ }\text{. }\frac{\sqrt{3}}{2}}{2\sqrt{3}}=\frac{1}{2}\Rightarrow A=\frac{\pi }{6}\] and \[\sin B=\frac{b\sin C}{c}=\frac{4\text{ }\text{. }\frac{\sqrt{3}}{2}}{2\sqrt{3}}=1\] Þ \[B=\frac{\pi }{2}\].
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