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 $\Delta A B C, B=\tan ^{-1}\left(\frac{1}{2}\right)$
$C=\tan ^{-1}\left(\frac{1}{3}\right)$
We know that in $\Delta A B C$,
$A+B+C=\pi$
$\Rightarrow A=\pi-\tan ^{-1}\left(\frac{1}{2}\right)-\tan ^{-1}\left(\frac{1}{3}\right)$
$\Rightarrow A=\pi-\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)$
$\Rightarrow A=\pi-\tan ^{-1}\left(\frac{5 / 6}{5 / 6}\right)=\pi-\tan ^{-1}(1)$
$\Rightarrow A=\pi-\tan ^{-1}\left(\tan \frac{\pi}{4}\right)$
$\Rightarrow A=\pi-\frac{\pi}{4} \Rightarrow A=\frac{3 \pi}{4}$
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, $(A, b)=\frac{3 \pi}{4}, \frac{2 \sqrt{2}}{\sqrt{5}}$