Q92 / 180 · KCET 2010 · Mathematics
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
- A$\frac{3 \pi}{4}, \frac{2}{\sqrt{5}}$
- B$\frac{\pi}{4}, \frac{2 \sqrt{2}}{\sqrt{5}}$
- Correct answerC$\frac{3 \pi}{4}, \frac{2 \sqrt{2}}{\sqrt{5}}$
- D$\frac{\pi}{4}, \frac{2}{\sqrt{5}}$
Solution
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, $(A, b)=\frac{3 \pi}{4}, \frac{2 \sqrt{2}}{\sqrt{5}}$
Hence, the correct answer is option: 'C'.