A) \[\frac{3\pi }{4},\frac{2}{\sqrt{5}}\]
B) \[\frac{\pi }{4},\frac{2\sqrt{2}}{\sqrt{5}}\]
C) \[\frac{3\pi }{4},\frac{2\sqrt{2}}{\sqrt{5}}\]
D) \[\frac{\pi }{4},\frac{2}{\sqrt{5}}\]
Correct Answer: C
Solution :
Given that \[a=2\] In \[\Delta ABC,\] \[B={{\tan }^{-1}}\left( \frac{1}{2} \right)\] \[C={{\tan }^{-1}}\left( \frac{1}{3} \right)\] We know that in \[\Delta ABC,\] \[A+B+C=\pi \] \[\Rightarrow \] \[A=\pi -B-C\] \[\Rightarrow \] \[A=\pi -{{\tan }^{-1}}\left( \frac{1}{2} \right)-{{\tan }^{-1}}\left( \frac{1}{3} \right)\] \[\Rightarrow \] \[A=\pi -{{\tan }^{-1}}\left( \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}} \right)\] \[\Rightarrow \] \[A=\pi -{{\tan }^{-1}}\left( \frac{5/6}{5/6} \right)=\pi -{{\tan }^{-1}}(1)\] \[\Rightarrow \] \[A=\pi -{{\tan }^{-1}}\left( \tan \frac{\pi }{4} \right)\] \[\Rightarrow \] \[A=\pi -\frac{\pi }{4}\Rightarrow A=\frac{3\pi }{4}\] Now, \[\sin A=\sin \frac{3\pi }{4}\] \[=\sin {{135}^{o}}=\cos {{45}^{o}}=\frac{1}{\sqrt{2}}\] \[\sin \,B=\frac{1}{\sqrt{3}}\] \[\left( \because \,\,\tan \,\,B=\frac{1}{2} \right)\] Now, by sine law \[\frac{a}{\sin A}=\frac{b}{\sin B}\] \[b=a.\frac{\sin B}{\sin A}=2.\frac{\frac{1}{\sqrt{5}}}{\frac{1}{\sqrt{2}}}=\frac{2\sqrt{2}}{\sqrt{5}}\] Hence, \[(A,b)=\frac{3\pi }{4},\frac{2\sqrt{2}}{\sqrt{5}}\]
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