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}+B\,given \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) $