Question

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

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

Answer 1

Answer: (B)

By sine rule $\frac{\sin A}{a}=\frac{\sin B}{b}$
$\Rightarrow$ $\frac{\sin \left(\frac{\pi }{2}+B\right)}{5}=\frac{\sin B}{4}$
$\left(\because A=\frac{\pi }{2}+Bgiven\right)$
$\Rightarrow$ $\tan B=\frac{4}{5}$
Also $\angle A+\angle B+\angle C=180{}^{\circ }$
$\Rightarrow$ $\frac{\pi }{2}+2\angle B+\angle C=\pi$
$\Rightarrow$ $2{\tan }^{-1}\left(\frac{4}{5}\right)+\angle C=\frac{\pi }{2}$
$\Rightarrow$ $\angle C=\frac{\pi }{2}-2{\tan }^{-1}\left(\frac{4}{5}\right)$
$\Rightarrow$ $\angle C=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{\frac{8}{5}}{1-\frac{16}{25}}\right)$
$=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{40}{9}\right)={\cot }^{-1}\left(\frac{40}{9}\right)$
$\Rightarrow$ $\angle C={\tan }^{-1}\left(\frac{9}{40}\right)$