Trigonometric Equations Question 298
Question: If the angles $ A,B,C $ of a triangle are in A.P. and the sides $ a,b,c $ opposite to these angles are in G. P. then $ a^{2},b^{2},c^{2} $ are in
[MP PET 1998]
Options:
A) A. P.
B) H. P.
C) G. P.
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- Since A, B and C are in A.P., therefore $ B=60{}^\circ $ and $ b^{2}=ac $ . $ \cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}\Rightarrow \frac{1}{2}=\frac{a^{2}+c^{2}-b^{2}}{2b^{2}} $ , $ (\because ,b^{2}=ac) $ $ b^{2}=a^{2}+c^{2}-b^{2}\Rightarrow a^{2}+c^{2}=2b^{2} $ .
BETA
