A) \[B{{r}^{-}}\]
B) \[HOCl\]
C) \[HCl{{O}_{2}}\]
D) \[HCl{{O}_{3}}\]
Correct Answer: A
Solution :
We have, a = 4, b = 3 and \[\angle A={{60}^{o}}\] \[\because \]\[\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\] \[\therefore \]\[\cos {{60}^{o}}=\frac{9+{{c}^{2}}-16}{6c}\] \[\Rightarrow \]\[\frac{1}{2}\times 6c={{c}^{2}}-7\]\[\Rightarrow \]\[{{c}^{2}}-3c-7=0\]
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