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

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]