Question

If in $∆\mathrm{ABC},a\mathit{=}\mathit{5}\mathit{,}\mathit{}b\mathit{=}\mathit{4}\mathit{,}\mathit{}A\mathit{=}\frac{\mathit{\pi }}{\mathit{2}}\mathit{+}B$, then $C$

• A. is ${\tan }^{-1}\left(\frac{1}{2}\right)$
• B. is ${\tan }^{-1}\left(\frac{9}{40}\right)$
• C. cannot be evaluated.
• D. is ${\tan }^{-1}\left(\frac{1}{9}\right)$

Answer 1

The correct option is B

is ${\tan }^{-1}\left(\frac{9}{40}\right)$

Explanation for the correct option.

Find the value of $C$:

Given,

$a\mathit{=}\mathit{5}\mathit{,}\mathit{}b\mathit{=}\mathit{4}\mathit{,}\mathit{}A\mathit{=}\frac{\pi }{\mathit{2}}\mathit{+}B$.

We know that,

$\begin{array}{rcl}\tan \left(\frac{A-B}{2}\right)& =& \frac{a-b}{a+b}cot\left(\frac{C}{2}\right)\\ & \Rightarrow & \tan \left(\frac{{\displaystyle \frac{\mathrm{\pi }}{2}}}{2}\right)=\frac{5-4}{5+4}cot\left(\frac{C}{2}\right)\\ & \Rightarrow & \frac{1}{cot{\displaystyle \frac{C}{2}}}=\frac{1}{9}\mathbf{[}\mathbf{\because }\mathbf{t}\mathbf{a}\mathbf{n}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{]}\\ & \Rightarrow & \tan \frac{C}{2}=\frac{1}{9}\mathbf{[}\mathbf{\because }\frac{\mathbf{1}}{\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{\theta }}\mathbf{=}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\theta }\mathbf{]}\end{array}$

Again, we know that,

$\begin{array}{rcl}\tan C& =& \frac{2\tan \left({\displaystyle \frac{C}{2}}\right)}{\left(1-{\tan }^2\left({\displaystyle \frac{C}{2}}\right)\right)}\\ & =& \frac{{\displaystyle \frac{2}{9}}}{1-{\displaystyle \frac{1}{81}}}\\ & =& \frac{9}{40}\end{array}$

Therefore, $C={\tan }^{-1}\begin{array}{rcl}& & \left(\frac{9}{40}\right)\end{array}$

Hence, the correct option B.