Sequence-And-Series Question 656
Question: If the angles A< B<C of a triangle are in A.P., then
Options:
A) $ c^{2}=a^{2}+b^{2}-ab $
B) $ b^{2}=a^{2}+c^{2}-ac $
C) $ c^{2}=a^{2}+b^{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ A+C=2B $ and $ A+B+C=180{}^\circ $ so, $ B=60{}^\circ $
$ \therefore ,\cos ,60{}^\circ =\frac{a^{2}+c^{2}-b^{2}}{2ac}\Rightarrow b^{2}=a^{2}+c^{2}-ac $
BETA
