Question

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

• A. $\left(\frac{3\pi }{4},\frac{2\sqrt{2}}{\sqrt{5}}\right)$
• B. $\left(\frac{\pi }{4},\frac{2}{\sqrt{5}}\right)$
• C. $\left(\frac{3\pi }{4},\frac{2}{\sqrt{5}}\right)$
• D. $\left(\frac{\pi }{4},\frac{2\sqrt{2}}{\sqrt{5}}\right)$

Answer 1

Answer: (A)

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 $\Delta ABC$,
$A+B+C=\pi$
$\Rightarrow A=\pi -{\tan }^{-1}\left(\frac{1}{2}\right)-{\tan }^{-1}\left(\frac{1}{3}\right)$
$\Rightarrow A=\pi -{\tan }^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)$
$\Rightarrow A=\pi -{\tan }^{-1}\left(\frac{5/6}{5/6}\right)=\pi -{\tan }^{-1}(1)$
$\Rightarrow A=\pi -{\tan }^{-1}\left(\tan \frac{\pi }{4}\right)$
$\Rightarrow A=\pi -\frac{\pi }{4}\Rightarrow A=\frac{3\pi }{4}$
Now, $\sin A=\sin \frac{3\pi }{4}$
$=\sin 135^{\circ }=\cos 45^{\circ }=\frac{1}{\sqrt{2}}$
$\sin B=\frac{1}{\sqrt{5}}$
$\left(\because \tan B=\frac{1}{2}\right)$
Now, by sine law
$\frac{a}{\sin A}=\frac{b}{\sin B}$
$b=a\cdot \frac{\sin B}{\sin A}=2\cdot \frac{\frac{1}{\sqrt{5}}}{\frac{1}{\sqrt{2}}}=\frac{2\sqrt{2}}{\sqrt{5}}$
Hence, $(A,b)=\frac{3\pi }{4},\frac{2\sqrt{2}}{\sqrt{5}}$