A) \[\frac{\pi }{3}\]
B) \[\frac{2\pi }{3}\]
C) \[\frac{\pi }{2}\]
D) \[\frac{3\pi }{4}\]
Correct Answer: B
Solution :
According to question \[R=A=B\] Since, \[{{R}^{2}}={{A}^{2}}+{{B}^{2}}+2AB\text{ }cos\theta \] \[\Rightarrow \] \[{{A}^{2}}={{A}^{2}}+{{A}^{2}}+2{{A}^{2}}\cos \theta \] \[\Rightarrow \] \[{{A}^{2}}=2{{A}^{2}}(1+\cos \theta )\] \[\Rightarrow \] \[1=2(1+\cos \theta )\] \[\Rightarrow \] \[\cos \theta =-\frac{1}{2}\] \[\Rightarrow \] \[\cos \theta =\cos 120{}^\circ \] \[\Rightarrow \] \[\theta =120{}^\circ \]or \[\frac{2\pi }{3}\]
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