A) 0.5
B) 0.75
C) 0.33
D) \[\frac{1}{\sqrt{3}}\]
Correct Answer: A
Solution :
[a] \[\cos A\sin B=\sin C\Rightarrow \sin (A+B)-sin(A-B)=2sinC\]\[\Rightarrow \sin C=\sin (B-A)\Rightarrow A+C=B\] \[(\because \,\,\,A+B=\pi -C)\] \[\therefore \,\,\,\,\,B=\frac{\pi }{2}\] Now \[3b-5c=0\Rightarrow 3-5\sin C=0\] \[\therefore \,\,\,\,\sin C=\frac{3}{5}\,\,and\,\,A=\frac{\pi }{2}-C\] \[\Rightarrow \cos A=\sin C\Rightarrow \frac{1-{{\tan }^{2}}\frac{A}{2}}{1+{{\tan }^{2}}\frac{A}{2}}=\frac{3}{5}\] \[\therefore \,\,\,\,\,\,{{\tan }^{2}}\frac{A}{2}=\frac{1}{4}\Rightarrow \tan \frac{A}{2}=0.5\]
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