Trigonometric Equations Question 194
Question: In a $ \Delta ABC $ , if $ c^{2}+a^{2}-b^{2}=ac $ , then $ \angle B= $
[MP PET 1983, 89, 90]
Options:
A) $ \frac{\pi }{6} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{3} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \cos B=\frac{c^{2}+a^{2}-b^{2}}{2ac}\Rightarrow \cos B=\frac{1}{2}i.e.,B=\frac{\pi }{3} $ .
BETA
