Answer:
see explanation
Step-by-step explanation:
Since the triangle is right use cosine or sine to solve for AC
note cos30° = [tex]\frac{\sqrt{3} }{2}[/tex], thus
cos30° = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{AC}{BC}[/tex] =[tex]\frac{AC}{\frac{10}{3} }[/tex] and
[tex]\frac{AC}{\frac{10}{3} }[/tex] = [tex]\frac{\sqrt{3} }{2}[/tex], that is
[tex]\frac{3x}{10}[/tex] = [tex]\frac{\sqrt{3} }{2}[/tex] ( cross- multiply )
6x = 10[tex]\sqrt{3}[/tex] ( divide both sides by 6 )
x = AC = [tex]\frac{10}{6}[/tex] [tex]\sqrt{3}[/tex] = [tex]\frac{5}{3}[/tex] [tex]\sqrt{3}[/tex]
