In \(\triangle \mathrm{ABC}\), if \(\mathrm{a}=2, \mathrm{~B}=\tan ^{-1} \frac{1}{2}\) and \(C=\tan ^{-1} \frac{1}{3}\), then (A, b) equals

Given that, \(\mathrm{a}=2\)
In \(\triangle \mathrm{ABC}, \mathrm{B}=\tan ^{-1}\left(\frac{1}{2}\right)\)
\[
C=\tan ^{-1}\left(\frac{1}{3}\right)
\]
We know that in \(\triangle \mathrm{ABC}\),
\(A+B+C=\pi\)
\(\Rightarrow \quad \mathrm{A}=\pi-\mathrm{B}-\mathrm{C}\)
\(\Rightarrow \quad \mathrm{A}=\pi-\tan ^{-1}\left(\frac{1}{2}\right)-\tan ^{-1}\left(\frac{1}{3}\right)\)
\(\Rightarrow \quad \mathrm{A}=\pi-\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)\)
\(\Rightarrow \quad \mathrm{A}=\pi-\tan ^{-1}\left(\frac{5 / 6}{5 / 6}\right)=\pi-\tan ^{-1}(1)\)
\(\Rightarrow \quad \mathrm{A}=\pi-\tan ^{-1}\left(\tan \frac{\pi}{4}\right)\)
\(\Rightarrow \quad \mathrm{A}=\pi-\frac{\pi}{4} \Rightarrow \mathrm{A}=\frac{3 \pi}{4}\)
Now, \(\quad \sin A=\sin \frac{3 \pi}{4}\)
\[
\begin{aligned}
=& \sin 135^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}} \\
\sin \mathrm{B} &=\frac{1}{\sqrt{5}} \quad\left(\because \tan \mathrm{B}=\frac{1}{2}\right)
\end{aligned}
\]
Now, by sine law
\[
\begin{gathered}
\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}} \\
\mathrm{b}=\mathrm{a} \cdot \frac{\sin \mathrm{B}}{\sin \mathrm{A}}=2 \cdot \frac{\frac{1}{\sqrt{5}}}{\frac{1}{\sqrt{2}}}=\frac{2 \sqrt{2}}{\sqrt{5}}
\end{gathered}
\]
Hence, \((A, b)=\frac{3 \pi}{4}, \frac{2 \sqrt{2}}{\sqrt{5}}\)