Q66 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics

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 $\angle A=$

  • A
    $60^{\circ}$
  • B
    $30^{\circ}$
  • C
    $150^{\circ}$
  • Correct answer
    D
    $120^{\circ}$

Solution

Sine rule is $$ \Rightarrow \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k $$ Then $\sin \mathrm{A}=\mathrm{ak}, \sin \mathrm{B}=\mathrm{bk}, \sin \mathrm{C}=\mathrm{ck}$ $$ \begin{aligned} & \text { Now, }(\sin A+\sin B)(\sin A-\sin B)=\sin C(\sin B+\sin C) \\ & \Rightarrow(a k+b k)(a k-b k)=k c(k b+k c) \\ & \Rightarrow a^2-b^2=b c+c^2 \Rightarrow a^2-b^2-c^2=b c \\ & \Rightarrow \frac{a^2-b^2-c^2}{2 b c}=\frac{1}{2} \Rightarrow \frac{b^2+c^2-a^2}{2 b c}=-\frac{1}{2} \\ & \Rightarrow \cos A=-\frac{1}{2}=\cos \frac{2 \pi}{3} \Rightarrow A=\frac{2 \pi}{3}=120^{\circ} \end{aligned} $$

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

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