Q66 / 160 · AP EAMCET 2022 (08 Jul Shift 1) · Mathematics

In a triangle $\mathrm{ABC},(\mathrm{b}+\mathrm{c}) \sin \frac{A}{2}=$

  • A
    $a \cos A$
  • Correct answer
    B
    $a \cos \left(\frac{B-C}{2}\right)$
  • C
    $a \sin \left(\frac{B+C}{2}\right)$
  • D
    $a \sin \left(\frac{B-C}{2}\right)$

Solution

In $\varnothing \mathrm{ABC}(\mathrm{b}+\mathrm{c}) \operatorname{Sin}\left(\frac{\mathrm{A}}{2}\right)=$ ? By $\operatorname{Sin}$ rule $\frac{a}{\operatorname{Sin} A}=\frac{b}{\operatorname{Sin} A}=\frac{c}{\operatorname{Sin} C}=k$ $\Rightarrow \mathrm{a}=\mathrm{k} \sin \mathrm{A}, \mathrm{b}=\mathrm{k} \sin \mathrm{B}, \mathrm{c}=\mathrm{k} \sin \mathrm{C}$ Now $(b+c) \sin \left(\frac{A}{2}\right)=(K \sin B+K \sin C) \operatorname{Sin}\left(\frac{A}{2}\right)$ $=\mathrm{k}(\sin \mathrm{B}+\sin \mathrm{C}) \cdot \operatorname{Sin}\left(\frac{\mathrm{A}}{2}\right)$ $\begin{aligned} & =\mathrm{k}\left[2 \sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right) \cdot \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)\right] \sin \left(\frac{\mathrm{A}}{2}\right) \\ & =\mathrm{k}\left[2 \sin \left(\frac{\pi+\mathrm{A}}{2}\right) \cdot \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)\right] \sin \left(\frac{\mathrm{A}}{2}\right) \\ & =\mathrm{k}\left[2 \sin \left(\frac{\pi}{2}-\frac{\mathrm{A}}{2}\right) \cdot \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)\right] \sin \left(\frac{\mathrm{A}}{2}\right) \\ & =\mathrm{k}\left[2 \sin \left(\frac{\mathrm{A}}{2}\right) \cdot \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)\right] \sin \left(\frac{\mathrm{A}}{2}\right) \\ & =\mathrm{k} \operatorname{Sin} \mathrm{A} \cdot \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)\end{aligned}$ $=\mathrm{a} \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)$

Hence, the correct answer is option: 'B'.

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