A) \[1/3\]
B) \[1/9\]
C) \[1/27\]
D) \[~1/18\]
Correct Answer: B
Solution :
Given, \[4\,\,\sin \,A=4\,\sin B=3\,\sin C\] or \[\frac{\sin A}{1/4}=\frac{\sin B}{1/4}=\frac{\sin C}{1/3}\] or \[\frac{\sin A}{3}=\frac{\sin B}{3}=\frac{\sin C}{4}\] Here, \[a=3k,b=3k\] and \[c=4k\] \[\therefore \] \[\cos \,C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\] \[=\frac{9{{k}^{2}}+9{{k}^{2}}-16{{k}^{2}}}{2\times 3k\times 3k}=\frac{1}{9}\]
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