Question

If $ 4\,\sin \,A=4\,\sin B=3\,\sin \,C $ in a triangle $ABC$, then $ \cos \,\,C $ is equal to

• A. $ \frac{1}{3} $
• B. $ \frac{1}{9} $
• C. $ \frac{1}{27} $
• D. $ \frac{1}{18} $

Answer 1

Answer: (B)

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} $