Question

If $\sin ({120}^0-\alpha )=\sin ({120}^0-\beta ),0<\alpha ,\beta <\pi$, then find the relation between $\alpha$ and $\beta$

• A. $\alpha =\beta$
• B. $\alpha +\beta ={60}^0$
• C. $\alpha +\beta ={90}^0$
• D. $\alpha =2\beta$

Answer 1

Answer: (A), (B)

The correct options are
A $\alpha =\beta$
B $\alpha +\beta ={60}^0$

If $\sin A=\sin B$, where $A={120}^0-\alpha$ and $B={120}^0-\beta$, then

$A=B$ or $A=\pi -B$, i.e., $A+B=\pi$

$\Rightarrow$ ${120}^0-\alpha ={120}^0-\beta$, or ${120}^0-\alpha +{120}^0-\beta ={180}^0$

$\Rightarrow \alpha =\beta$ or $\alpha +\beta ={60}^0$