In any triangle $\mathrm{ABC}, \sin \frac{A}{2} \leq$

In any triangle $\mathrm{ABC}, \sin \frac{A}{2} \leq$
  1. $\frac{2 a}{\sqrt{b c}}$
  2. $\frac{a}{2 \sqrt{b c}}$
  3. $\frac{3 a}{\sqrt{b c}}$
  4. $\frac{\sqrt{b c}}{2 a}$

Solution

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ $\Rightarrow \mathrm{a}=\mathrm{k} \sin \mathrm{A}, \mathrm{b}=\mathrm{k} \sin \mathrm{B}, \mathrm{c}=\mathrm{k} \sin \mathrm{C}$ $\therefore \frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}=\frac{\sin \mathrm{A}}{\sin \mathrm{B}+\sin \mathrm{C}}=\frac{2 \sin \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{A}}{2}}{2 \sin \frac{\mathrm{B}+\mathrm{C}}{2}+\cos \frac{\mathrm{B}-\mathrm{C}}{2}}$ $\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}=\frac{\sin \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{A}}{2}}{\cos \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{B}-\mathrm{C}}{2}}$ $\therefore \sin \frac{\mathrm{A}}{2}=\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}} \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)$ $\because \frac{\mathrm{B}-\mathrm{C}}{2}$ in acuite $\therefore \quad \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right) 2 \leq 1$ $\therefore \sin \frac{A}{2} \leq \frac{a}{b+c}$ ...(i) We have $\mathrm{GM} \leq \mathrm{AM}$ $\begin{aligned} & 2 \sqrt{b c} \leq b+c \\ & \Rightarrow \frac{1}{b+c} \leq \frac{1}{2 \sqrt{b c}}\end{aligned}$ from (i) $\sin \frac{\mathrm{A}}{2} " \frac{\mathrm{a}}{2 \sqrt{\mathrm{bc}}}$

Asked in: AP EAMCET 2022 (08 Jul Shift 1)

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