A) \[\theta =\frac{2\pi }{3}\]
B) \[\theta =\frac{\pi }{2}\]
C) \[\theta =\pi \]
D) \[\theta =0\]
Correct Answer: B
Solution :
Given, \[A+B=C\] \[\Rightarrow \] \[{{C}^{2}}={{A}^{2}}+{{B}^{2}}+2AB\,cos\theta \] ...(i) But \[{{C}^{2}}={{A}^{2}}+{{B}^{2}}\] (given) .... (ii) Comparing Eqs. (i) and (ii), we get \[2AB\text{ }cos\theta =0\] or \[cos\theta =0\] \[\Rightarrow \] \[\theta =\frac{\pi }{2}\]
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