A) \[{{30}^{o}},{{60}^{o}},{{90}^{o}}\]
B) \[{{45}^{o}},{{60}^{o}},{{75}^{o}}\]
C) \[{{45}^{o}},{{45}^{o}},{{90}^{o}}\]
D) \[{{60}^{o}},{{60}^{o}},{{60}^{o}}\]
Correct Answer: B
Solution :
Since A,B,C are in A.P., therefore \[B={{60}^{o}}\][\[\because A+B+C=180\]and \[A+C=2B\]] Now, \[\sin (2A+B)=\frac{1}{2}\](given) Þ \[2A+B={{30}^{o}}\]or \[{{150}^{o}}\] But as \[B={{60}^{o}},\] \[2A+B\ne {{30}^{o}}\]. Hence \[2A+B={{150}^{o}}\Rightarrow A={{45}^{o}}\] Hence\[A={{45}^{o}}\], \[B={{60}^{o}},C={{75}^{o}}\].
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