Question

The sides a,b,c of $\Delta ABC$ are in G.P, and $\log a-\log 2b,\log 2b-\log 3c,log3c-\mathrm{loga}$ are in A.P., then $\Delta ABC$ is_____ angled triangle.

Answer 1

Given that, In $\Delta ABC$

Sides $a,b,c$ in G.P.

Then,

$\frac{b}{a}=\frac{c}{b}$ …….. (1)

$b^2=ac$ …… (2)

Now given that,

$\log a-\log 2b,\log 2b-\log 3c,log3c-\mathrm{loga}$ in A.P.

Then, we know that In an A.P.

$(\log 2b-\log 3c)-(\mathrm{loga}-\log 2b)=(\log 3c-\mathrm{loga})-(\log 2b-\log 3c)$

$\log 2b+\log 2b-\log 3c-\log a=\log 3c+\log 3c-\mathrm{loga}-\log 2b$

$2\log 2b-\log 3c-\log a=2\log 3c-\mathrm{loga}-\log 2b$

$2\log 2b+\log 2b=\log 3c+2\log 3c$

$3\log 2b=3\log 3c$

$\log 2b=\log 3c$

Eliminating $\log$ both side and we get,

$2b=3c$

$\frac{c}{b}=\frac{2}{3}$

By equation (1)

$\frac{b}{a}=\frac{c}{b}=\frac{2}{3}$

Then,

$b=\frac{2}{3}a$ and $c=\frac{2}{3}b$

By equation (2),

$b^2=ac$ put $b=\frac{2}{3}a$ and we get

${\left(\frac{2a}{3}\right)}^2=ac$

$c=\frac{4a}{9}$

Now, $a,b=\frac{2a}{3},c=\frac{4a}{9}$

Using $\cos$ rule and we gat

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\cos A=\frac{{\left(\frac{2a}{3}\right)}^2+{\left(\frac{4a}{9}\right)}^2-a^2}{2\times \frac{2a}{3}\times \frac{4a}{9}}$

$\cos A=\frac{{\frac{4a}{9}}^2+{\frac{16a}{81}}^2-a^2}{2\times \frac{2a}{3}\times \frac{4a}{9}}$

$\cos A=\frac{-29}{48}=-\frac{1}{2}$

$\cos A=\cos (-{45}^0)$

$cosA=\cos {135}^0$

And $A={135}^0$

Which is obtuse.