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\[2{{\tan }^{-1}}\left( \frac{1}{40} \right)\]
Correct Answer: B
Solution :
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 \[\angle A+\angle B+\angle C=180{}^\circ \] \[\Rightarrow \] \[\frac{\pi }{2}+2\angle B+\angle C=\pi \] \[\Rightarrow \] \[2{{\tan }^{-1}}\left( \frac{4}{5} \right)+\angle C=\frac{\pi }{2}\] \[\Rightarrow \] \[\angle C=\frac{\pi }{2}-2{{\tan }^{-1}}\left( \frac{4}{5} \right)\] \[\Rightarrow \] \[\angle C=\frac{\pi }{2}-{{\tan }^{-1}}\left( \frac{\frac{8}{5}}{1-\frac{16}{25}} \right)\] \[=\frac{\pi }{2}-{{\tan }^{-1}}\left( \frac{40}{9} \right)={{\cot }^{-1}}\left( \frac{40}{9} \right)\] \[\Rightarrow \] \[\angle C={{\tan }^{-1}}\left( \frac{9}{40} \right)\]
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