A) \[\frac{4}{5}\]
B) \[\frac{3}{20}\]
C) \[\frac{24}{25}\]
D) \[\frac{1}{50}\]
Correct Answer: C
Solution :
We know that, \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] \[=\frac{{{(3)}^{2}}+{{(5)}^{2}}-{{(4)}^{2}}}{2\times 3\times 5}\] \[=\frac{9+25-16}{30}=\frac{18}{30}=\frac{3}{5}\] \[\Rightarrow \] \[\sin \,B=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\] \[\therefore \] \[\sin \,2B=2\,\sin \,B\,\,cos\,B\] \[=2\times \frac{4}{5}\times \frac{3}{5}=\frac{24}{25}\]
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