Question

Help me to find AC please

Answer 1

Answer:

see explanation

Step-by-step explanation:

Since the triangle is right use cosine or sine to solve for AC

note cos30° = $\frac{\sqrt{3} }{2}$, thus

cos30° =  $\frac{adjacent}{hypotenuse}$ = $\frac{AC}{BC}$  =$\frac{AC}{\frac{10}{3} }$ and

$\frac{AC}{\frac{10}{3} }$ = $\frac{\sqrt{3} }{2}$, that is

$\frac{3x}{10}$ = $\frac{\sqrt{3} }{2}$ ( cross- multiply )

6x = 10$\sqrt{3}$ ( divide both sides by 6 )

x = AC = $\frac{10}{6}$ $\sqrt{3}$ = $\frac{5}{3}$ $\sqrt{3}$