Question

In a triangle a cos ^2(c/2)+c cos^(a/2)=3b/2 then its sides a,b,c are in

Answer 1

@kropot72 ,@Haseeb96

Answer 2

c^2 = a^2 + b^2 - 2ab cos(C)

b^2 = a^2 + c^2 - 2ac cos(B)

a^2 = b^2 + c^2 - 2bc cos(A)

Answer 3

\[\begin{array} \\
a \cos ^2(C/2)+c \cos^2(A/2) &=3b/2 \\
2a \cos ^2(C/2)+2c \cos^2(A/2) &=3b \\

a (1+\cos C)+c(1+\cos A) &=3b ~~~\color{gray}{\because 2\cos^2\theta /2 = 1+\cos\theta }\\
a + c + (a\cos C + c\cos A )&=3b \\

\end{array}\]

Answer 4

see if that looks okay so far ^^

Answer 5

ok

Answer 6

b = a cos C +c cos A

Answer 7

a+c+b=3b

Answer 8

a+c=2b

Answer 9

this is in h.p or g.p or a.p form

Answer 10

Excellent !

Answer 11

ofcourse it is AP

Answer 12

if a,b,c are in AP, then 2b = a+c

Answer 13

yes