A) \[-1\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{3}\]
D) \[\frac{2}{3}\]
E) 1
Correct Answer: A
Solution :
\[\because \]\[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] As\[a\to 2c\]and\[b\to 3c\] \[\therefore \]\[\cos B=\frac{4{{c}^{2}}+{{c}^{2}}-9{{c}^{2}}}{4{{c}^{2}}}=-\frac{4}{4}=-1\] When\[a\to 2c\]and\[b\to 3c,\]then\[cos\text{ }B\to -1.\]
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