Trigonometric Equations Question 209
Question: In $ \Delta ABC, $ if $ (a+b+c)(a-b+c) $ =3ac, then
[AMU 1996]
Options:
A) $ \angle B=60^{o} $
B) $ \angle B=30^{o} $
C) $ \angle C=60^{o} $
D) $ \angle A+\angle C=90^{o} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ {{(a+c)}^{2}}-b^{2}=3ac\Rightarrow a^{2}+c^{2}-b^{2}=ac $ But $ \cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}=\frac{1}{2}\Rightarrow B=\frac{\pi }{3} $ .
BETA
