Trigonometric Equations Question 326
Question: If in a triangle $ ABC,a=5,b=4,A=\frac{\pi }{2}+B $ , then C
[Kerala (Engg.) 2005]
Options:
A) Is $ {{\tan }^{-1}}( \frac{1}{9} ) $
B) Is $ {{\tan }^{-1}}\frac{1}{40} $
C) Cannot be evaluated
D) Is $ 2{{\tan }^{-1}}( 1/9 ) $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ \because $ $ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} $ ; $ \frac{5}{\sin ( \frac{\pi }{2}+B )}=\frac{4}{\sin B} $ $ \frac{5}{\cos B}=\frac{4}{\sin B} $ , \ $ \tan B=\frac{4}{5} $ $ \tan A=\tan ( \frac{\pi }{2}+B )=-\cot B=\frac{-5}{4} $ $ \tan C=\tan (\pi -(A+B)) $ $ =-\tan (A+B), $ $ [A+B+C=\pi ] $ $ =-\frac{(\tan A+\tan B)}{1-\tan A.\tan B} $ $ =\frac{-( -\frac{5}{4}+\frac{4}{5} )}{1+1}=\frac{9}{40} $ $ C={{\tan }^{-1}}( \frac{( 2.\frac{1}{9} )}{1-{{( \frac{1}{9} )}^{2}}} ) $ ; \ $ C=2{{\tan }^{-1}}( \frac{1}{9} ) $ .
BETA
