The sides a,b,c, of ΔABC are in G.P, and loga - log2b, log2b - log3c, log3c - loga are in A.P., then ΔABC
Given that, In ΔABC
Sides a,b,c in G.P.
Then,
ba=cb …….. (1)
b2=ac …… (2)
Now given that,
loga−log2b,log2b−log3c,log3c−loga in A.P.
Then, we know that In an A.P.
(log2b−log3c)−(loga−log2b)=(log3c−loga)−(log2b−log3c)
log2b+log2b−log3c−loga=log3c+log3c−loga−log2b
2log2b−log3c−loga=2log3c−loga−log2b
2log2b+log2b=log3c+2log3c
3log2b=3log3c
log2b=log3c
Eliminating log both side and we get,
2b=3c
cb=23
By equation (1)
ba=cb=23
Then,
b=23a and c=23b
By equation (2),
b2=ac put b=23a and we get
(2a3)2=ac
c=4a9
Now, a,b=2a3,c=4a9
Using cos rule and we gat
cosA=b2+c2−a22bc
cosA=(2a3)2+(4a9)2−a22×2a3×4a9
cosA=4a92+16a812−a22×2a3×4a9
cosA=−2948=−12
cosA=cos(−450)
cosA=cos1350
And A=1350