Trigonometric Equations Question 342
Question: In triangle $ ABC $ if $ A+C=2B $ , then $ \frac{a+c}{\sqrt{a^{2}-ac+c^{2}}} $ is equal to
[UPSEAT 1999]
Options:
A) $ 2\cos \frac{A-C}{2} $
B) $ \sin \frac{A+C}{2} $
C) $ \sin \frac{A}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- $ A+C=2B\Rightarrow B=60^{o} $ , $ \cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac} $ Since $ B=60^{o} $
$ \Rightarrow ac=a^{2}+c^{2}-b^{2} $
Þ $ b^{2}=a^{2}+c^{2}-ac $ Therefore $ \frac{a+c}{\sqrt{a^{2}-ac+c^{2}}}=\frac{a+c}{b}=\frac{\sin A+\sin C}{\sin B} $ $ =\frac{2\sin \frac{A+C}{2}\cos \frac{A-C}{2}}{2\sin \frac{B}{2}\sin \frac{A+C}{2}}=\frac{\cos \frac{A-C}{2}}{\sin \frac{B}{2}} $ $ =\frac{\cos \frac{A-C}{2}}{\sin 30^{o}}\Rightarrow 2\cos \frac{A-C}{2} $ .
BETA
