Trigonometric Equations Question 367
Question: If the median of $ \Delta ABC $ through A is perpendicular to $ AB $ , then
Options:
A) $ \tan A+\tan B=0 $
B) $ 2\tan A+\tan B=0 $
C) $ \tan A+2\tan B=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- We have $ BD=DC $ and $ \angle DAB=90^{o} $ . Draw CN perpendicular to BA produced, then in $ \Delta BCN $ , we have $ DA=\frac{1}{2}CN $ and $ AB=AN $ Let $ \angle CAN=\alpha $ $ \because \tan A=\tan (\pi -\alpha )=-\tan \alpha $ $ =-\frac{CN}{NA}=-2\frac{AD}{AB}=-2\tan B $
Þ $ \tan A+2\tan B=0 $ .
BETA
