Question

If $a,b,c$ be the sides of triangle $ABC$ and if roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal, then
${\sin }^2\left(\frac{A}{2}\right),{\sin }^2\left(\frac{B}{2}\right),{\sin }^2\left(\frac{C}{2}\right)$ are in

• A. A.P
• B. G.P
• C. H.P
• D. A.G.P

Answer 1

The correct option is C H.P

$\because a(b-c)+b(c-a)+c(a-b)=0$
$\therefore x=1$ is a root of the equation.
$a(b-c)x^2+b(c-a)x+c(a-b)=0$
Then the other root $=1$
$\because$roots are equal.
$\because \left|x\right|=\frac{c(a-b)}{a(b-c)}$
$\Rightarrow ab-ac=ca-cb$
$\Rightarrow b=\frac{2ac}{a+c}$
$\therefore a,b,c$ are in H.P.
$\Rightarrow \frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P
$\Rightarrow \frac{s}{a},\frac{s}{b},\frac{s}{c}$ are in A.P
$\Rightarrow \frac{s}{a}-1,\frac{s}{b}-1,\frac{s}{c}-1$ are in A.P
$\Rightarrow \frac{s-a}{a},\frac{s-b}{b},\frac{s-c}{c}$ are in A.P
Multiplying in each by $\frac{abc}{(s-a)(s-b)(s-c)}$ then
$\Rightarrow \frac{bc}{(s-b)(s-c)},\frac{ca}{(s-a)(s-c)},\frac{ab}{(s-a)(s-b)}$ are in A.P
$\Rightarrow \frac{(s-b)(s-c)}{bc},\frac{(s-a)(s-c)}{ca},\frac{(s-a)(s-b)}{ab}$ are in H.P
$\Rightarrow {\sin }^2\left(\frac{A}{2}\right),{\sin }^2\left(\frac{B}{2}\right),{\sin }^2\left(\frac{C}{2}\right)$ are in H.P.