Respuesta :

Answer:

b

Step-by-step explanation:

sinA = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{BC}{AB}[/tex] = [tex]\frac{s\sqrt{3} }{2s}[/tex] ( cancel s on numerator/ denominator ), then

sinA = [tex]\frac{\sqrt{3} }{2}[/tex] → b

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