Question

In a $∆ABC$, if ${\sin }^2\left(\frac{A}{2}\right),{\sin }^2\left(\frac{B}{2}\right)$ and ${\sin }^2\left(\frac{C}{2}\right)$ are in $\mathrm{HP}$. Then, $a,b$ and $c$ will be in

• A. $\mathrm{AP}$
• B. $\mathrm{GP}$
• C. $\mathrm{HP}$
• D. None of these

Answer 1

The correct option is C

$\mathrm{HP}$

Explanation for correct option

Given ${\sin }^2\left(\frac{A}{2}\right),{\sin }^2\left(\frac{B}{2}\right)$ and ${\sin }^2\left(\frac{C}{2}\right)$ are in $\mathrm{HP}$.

We know that,

$$\begin{gathered} \sin \left(\frac{A}{2}\right)=\sqrt{\frac{\left(s-b\right)\left(s-c\right)}{bc}} \\ \Rightarrow {\sin }^2\left(\frac{A}{2}\right)=\frac{\left(s-b\right)\left(s-c\right)}{bc} \end{gathered}$$

Similarly,

${\sin }^2\left(\frac{B}{2}\right)=\frac{\left(s-a\right)\left(s-c\right)}{ac}$ and ${\sin }^2\left(\frac{C}{2}\right)=\frac{\left(s-a\right)\left(s-b\right)}{ab}$

Now ${\sin }^2\left(\frac{A}{2}\right),{\sin }^2\left(\frac{B}{2}\right)$ and ${\sin }^2\left(\frac{C}{2}\right)$ are in $\mathrm{HP}$. Hence their reciprocal will be in $\mathrm{AP}$. Therefore,

$\frac{1}{{\sin }^2\left({\displaystyle \frac{A}{2}}\right)},\frac{1}{{\sin }^2\left({\displaystyle \frac{B}{2}}\right)}$ and $\frac{1}{{\sin }^2\left({\displaystyle \frac{C}{2}}\right)}$ are in $\mathrm{AP}$ which implies,

$\frac{bc}{\left(s-b\right)\left(s-c\right)},\frac{ac}{\left(s-a\right)\left(s-c\right)}$ and $\frac{ab}{\left(s-a\right)\left(s-b\right)}$ are in $\mathrm{AP}$.

Multiply and divide all terms by $s$ we get,

$\frac{sbc}{s\left(s-b\right)\left(s-c\right)},\frac{sac}{s\left(s-a\right)\left(s-c\right)}$ and $\frac{sab}{s\left(s-a\right)\left(s-b\right)}$ are in $\mathrm{AP}$.

Now divide and multiply first, second and third term by $\left(s-a\right),\left(s-b\right)$ and $\left(s-c\right)$ respectively, we get

$\frac{sbc\left(s-a\right)}{s\left(s-a\right)\left(s-b\right)\left(s-c\right)},\frac{sac\left(s-b\right)}{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ and $\frac{sab\left(s-c\right)}{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ are in $\mathrm{AP}$.

$\frac{sbc\left(s-a\right)}{{∆}^2},\frac{sac\left(s-b\right)}{{∆}^2}$ and $\frac{sab\left(s-c\right)}{{∆}^2}$ are in $\mathrm{AP}$. $\left[\because ∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\right]$

Now, as the given terms are in $\mathrm{AP}$,

$$\begin{gathered} \frac{2s\left(s-b\right)ac}{{∆}^2}=\frac{s\left(s-a\right)bc}{{∆}^2}+\frac{s\left(s-c\right)ab}{{∆}^2} \\ \Rightarrow 2\left(s-b\right)ac=\left(s-a\right)bc+\left(s-c\right)ab \\ \Rightarrow 2acs-2bca=bcs-abc+abs-abc \\ \Rightarrow 2acs=bcs+abs \\ \Rightarrow 2acs=s\left(bc+ab\right) \\ \Rightarrow 2ac=bc+ab \\ \Rightarrow b=\frac{2ac}{a+c} \end{gathered}$$

Therefore , $a,b$ and $c$ will be in $\mathrm{HP}$.

Hence, the correct answer is option (C).