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In $\triangle \mathrm{ABC}$, if $(\sin \mathrm{A}+\sin \mathrm{B})(\sin \mathrm{A}-\sin \mathrm{B})=\sin \mathrm{C}(\sin \mathrm{B}$ $+\sin C$ ), then $\angle A=$
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Verified Answer
The correct answer is:
$120^{\circ}$
Sine rule is
$$
\Rightarrow \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k
$$
Then $\sin \mathrm{A}=\mathrm{ak}, \sin \mathrm{B}=\mathrm{bk}, \sin \mathrm{C}=\mathrm{ck}$
$$
\begin{aligned}
& \text { Now, }(\sin A+\sin B)(\sin A-\sin B)=\sin C(\sin B+\sin C) \\
& \Rightarrow(a k+b k)(a k-b k)=k c(k b+k c) \\
& \Rightarrow a^2-b^2=b c+c^2 \Rightarrow a^2-b^2-c^2=b c \\
& \Rightarrow \frac{a^2-b^2-c^2}{2 b c}=\frac{1}{2} \Rightarrow \frac{b^2+c^2-a^2}{2 b c}=-\frac{1}{2} \\
& \Rightarrow \cos A=-\frac{1}{2}=\cos \frac{2 \pi}{3} \Rightarrow A=\frac{2 \pi}{3}=120^{\circ}
\end{aligned}
$$
$$
\Rightarrow \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k
$$
Then $\sin \mathrm{A}=\mathrm{ak}, \sin \mathrm{B}=\mathrm{bk}, \sin \mathrm{C}=\mathrm{ck}$
$$
\begin{aligned}
& \text { Now, }(\sin A+\sin B)(\sin A-\sin B)=\sin C(\sin B+\sin C) \\
& \Rightarrow(a k+b k)(a k-b k)=k c(k b+k c) \\
& \Rightarrow a^2-b^2=b c+c^2 \Rightarrow a^2-b^2-c^2=b c \\
& \Rightarrow \frac{a^2-b^2-c^2}{2 b c}=\frac{1}{2} \Rightarrow \frac{b^2+c^2-a^2}{2 b c}=-\frac{1}{2} \\
& \Rightarrow \cos A=-\frac{1}{2}=\cos \frac{2 \pi}{3} \Rightarrow A=\frac{2 \pi}{3}=120^{\circ}
\end{aligned}
$$
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