Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $cosA+2cosB+cosC=2$ then $a,b,c$ are in?

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. None of the above

Answer 1

The correct option is A $A.P.$

$cosA+cosC=2(1-cosB)$

Or $2cos\left(\frac{A+C}{2}\right).cos\left(\frac{A-C}{2}\right)=4si{n}^2\frac{B}{2}$
$2sin\frac{B}{2}.cos\left(\frac{A-C}{2}\right)=4si{n}^2\frac{B}{2}$

Or, $sin\frac{B}{2}=0$

Or, $\frac{B}{2}=\pi$

$B=2\pi$ ...(i) not possible.

Then
$cos\left(\frac{A-C}{2}\right)=2sin\left(\frac{B}{2}\right)$
$\frac{cos\left(\frac{A-C}{2}\right)}{sin\left(\frac{B}{2}\right)}=2$

Or ,$\frac{a+c}{b}=2$

Or ,$a+c=2b$.

Hence the sides are in A.P.