BITSAT 2015 Paper

Let A, B and C be the three angles of ΔABC and
Let a = 10 and b = 9
It is given that the angles are in AP.
∴ 2B = A + C on adding B both the sides, we get
3B = A + B + C
⇒ 3B = 180° ⇒ B = 60°
Now, we know cos B =

a2+c2−b2

2ac

⇒ cos 60° =

102+c2−92

2×10×c

⇒

1

2

=

100+c2−81

20c

⇒ c2 - 10x + 19 = 0 ⇒ c = 5 ± √6