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If $\sin (120-A)=\sin (120-B)$ and $0 < A, B < \pi$ then all values of $A$ and $B$ are given by
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Verified Answer
The correct answer is:
$A=B$ or $A+B=\frac{\pi}{3}$
Given, equation is
$$
\sin (120-A)=\sin (120-B)
$$
Since, sine is positive in II quadrant.
$\therefore$ Either $120-A=120-B$
$\Rightarrow \quad A=B$
or $\quad 120-A=180-(120-B)$
$\Rightarrow \quad 120-A=60+B \Rightarrow A+B=\frac{\pi}{3}$
$$
\sin (120-A)=\sin (120-B)
$$
Since, sine is positive in II quadrant.
$\therefore$ Either $120-A=120-B$
$\Rightarrow \quad A=B$
or $\quad 120-A=180-(120-B)$
$\Rightarrow \quad 120-A=60+B \Rightarrow A+B=\frac{\pi}{3}$
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