Q66 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics
In a $\triangle A B C$, if $\angle A=60^{\circ}$, then $(a+b+c)(b+c-a)=$
- Correct answerA$3 b c$
- B$2 a b c$
- C$a b c$
- D$a+b+c$
Solution
In $\triangle A B C$,
$\angle A=60^{\circ}, \angle B+\angle C=180^{\circ}-\angle A=120^{\circ}$
$a=k \sin A, b=k \sin B, c=k \sin C$ (sine rule)
Now, $(b+c+a)(b+c-a)$
$\begin{aligned} & =(b+c)^2-a^2 \\ & =k^2(\sin B+\sin C)^2-k^2 \sin ^2 A \\ & =K^2\left(2 \sin \frac{B+C}{2} \cos \frac{B-C}{2}\right)^2-K^2\left(\frac{3}{4}\right) \\ & \quad\left[\because \sin A=\frac{\sqrt{3}}{2}\right]\end{aligned}$
$\begin{aligned} & =k^2\left(2 \sin 60^{\circ} \cdot \cos \frac{B-C}{2}\right)^2-\frac{3 k^2}{4} \\ & =3 k^2 \cos ^2 \frac{B-C}{2}-\frac{3 k^2}{4} \\ & =3 k^2\left(\cos ^2 \frac{B-C}{2}-\frac{1}{4}\right) \\ & =\frac{3 k^2}{2}\left(2 \cos ^2 \frac{B-C}{2}-1+\frac{1}{2}\right) \\ & =\frac{3 k^2}{2}\left(\cos (B-C)+\frac{1}{2}\right) \\ & =\frac{3 k^2}{2}(\cos (B-C)+\cos A) \\ & =\frac{3 k^2}{2}(\cos (B-C)-\cos (B+C)) \\ & =\frac{3 k^2}{2} \times 2 \sin B \times \sin C\end{aligned}$
$=3(k \sin B)(k \sin C)=3 b c$
Hence, the correct answer is option: 'A'.