Since,a,bandcare in AP So,2b=a+c Also,∠A=2∠C⋅⋅⋅⋅⋅⋅⋅(i) Now, in a△ABC, a=2RsinA,b=2RsinB,c=2RsinC So, 2b=a+c ⇒2⋅2RsinB=2RsinA+2RsinC ⇒2sinB=sinA+sinC In a △ABC, A+B+C=180∘ ⇒2C+B+C=180∘ ⇒B=180∘−3C So,sinB=sin(180∘−3C) =sin(3C)⋅⋅⋅⋅⋅⋅⋅(ii) And2sinB=sinA+sinC ⇒2sin(3C)=sin(2C)+sinC [using Eq. (i) and (ii)] =2sinCcosC+sinC and sin(3C)=3sinC−4sin3C So, 2sin(3C)=2(3−4sin2C) =2cosC+1 =2(4cos2C−1) ⇒8cos2C−2=2cosC+1 ⇒8cos2C−2cosC−3=0 ∴cosC=
2±√4−4(8)(−3)
16
=
2±√100
16
=
2±10
16
So,cosC=
3
4
,cosC=−
1
2
Since,Cmust be positive, So,cosC=
3
4
Now, cosA=cos(2C) =2cos2C−1=2(
3
4
)2−1=
1
8
And B=180∘−3C cosB=cos(180∘−3C) =−cos(3C) =−(4cos3C−3cosC) =−(4(
