Question

In $△ABC,$ if ${\sin }^2\frac{A}{2},{\sin }^2\frac{B}{2},$ and ${\sin }^2\frac{C}{2}$ are in H.P., then $a,b$ and $c$ will be in

• A. A . P.
• B. G . P.
• C. H . P.
• D. none of these

Answer 1

Answer: (C)

H . P.

since, they are in $HP$

$si{n}^2\frac{B}{2}=\frac{2si{n}^2\frac{A}{2}si{n}^2\frac{C}{2}}{si{n}^2\frac{A}{2}+si{n}^2\frac{C}{2}}\Rightarrow \frac{(s-a)(s-c)}{ac}=\frac{\frac{2(s-b)(s-c)(s-a)(s-b)}{a{b}^{2}c}}{\frac{(s-b)(s-c)}{bc}+\frac{(s-a)(s-b)}{ab}}$

After Simplifying ,

$\Rightarrow b(as+cs-2ca)=2ac(s-b)\Rightarrow sba+sbc=2acs\Rightarrow ba+bc=2ac\Rightarrow b=\frac{2ac}{a+c}$

Hence, a,b,c are in $HP$

Therefore, Answer is $C$