desiree05lopez desiree05lopez
  • 24-02-2021
  • Mathematics
contestada

I need help ASAP. Thank you. Special right triangle.

I need help ASAP Thank you Special right triangle class=

Respuesta :

jimrgrant1 jimrgrant1
  • 24-02-2021

Answer:

JK = 3[tex]\sqrt{3}[/tex]

Step-by-step explanation:

Using the cosine ratio in the right triangle and the exact value

cos60° = [tex]\frac{1}{2}[/tex] , then

cos60° = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{JK}{JL}[/tex] = [tex]\frac{JK}{6\sqrt{3} }[/tex] = [tex]\frac{1}{2}[/tex] ( cross- multiply )

2JK = 6[tex]\sqrt{3}[/tex] ( divide both sides by 2 )

JK = 3[tex]\sqrt{3}[/tex]

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