gfredmoore9254 gfredmoore9254

31-12-2020
Mathematics

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the question is below

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jimrgrant1 jimrgrant1

31-12-2020

Answer:

sin I = [tex]\frac{80}{89}[/tex]

Step-by-step explanation:

sin I = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{KJ}{JI}[/tex] = [tex]\frac{80}{89}[/tex]

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