Question

Plssss help I need AC

Answer 1

∠ACB=90° ⇒ ΔABC is right triangle

AB is hypotenuse, ∠BAC=30° ⇒

$CB=\dfrac{AB}{2}=\dfrac{12\sqrt{3} }{2} =6\sqrt{3}$

Let's use Pythagorean theorem

$AB^2=BC^2+AC^2\\AC^2=AB^2-BC^2\\AC^2=(12\sqrt{3})^2-(6\sqrt{3} )^2\\AC=\sqrt{(12\sqrt{3}-6\sqrt{3})(12\sqrt{3}+6\sqrt{3}))} \\AC=\sqrt{6\sqrt{3}\times18\sqrt{3} } \\AC=\sqrt{3\times6\times18} \\AC=18$

Answer 2

Answer:

AC = 18

Step-by-step explanation:

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

cos30° = $\frac{\sqrt{3} }{2}$ , then

cos30° = $\frac{adjacent}{hypotenuse}$ = $\frac{AC}{AB}$ = $\frac{AC}{12\sqrt{3} }$ = $\frac{\sqrt{3} }{2}$ ( cross- multiply )

2 AC = 12$\sqrt{3}$ × $\sqrt{3}$ = 12 × 3 = 36 ( divide both sides by 2 )

AC = 18