Trigonometric Equations Question 158
Question: If in a $ \Delta ABC $ , $ \cos 3A+\cos 3B+\cos 3C=1 $ , then one angle must be exactly equal to
Options:
A) $ 90^{o} $
B) $ 45^{o} $
C) $ 120^{o} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- Since $ \cos 3A+\cos 3B+\cos 3C=1 $
Þ $ 4\sin \frac{3A}{2}\sin \frac{3B}{2}\sin \frac{3C}{2}=0 $ Either $ \frac{3A}{2}=180^{o} $ or $ \frac{3B}{2}=180^{o} $ or $ \frac{3C}{2}=180^{o} $ Either $ A=120^{o} $ or $ B=120^{o} $ or $ C=120^{o} $ .
BETA
