Answer:
Part 1) [tex]x=18\ units[/tex]
Part 2) [tex]y=6\sqrt{3}\ units[/tex]
Step-by-step explanation:
step 1
Find the value of x
we know that
In the right triangle of the figure
[tex]cos(30^o)=\frac{x}{12\sqrt{3}}[/tex] ----> adjacent side divided by hypotenuse (CAH)
Remember that
[tex]cos(30^o)=\frac{\sqrt{3}}{2}[/tex]
substitute
[tex]\frac{\sqrt{3}}{2}=\frac{y}{12\sqrt{3}}[/tex]
[tex]x=18\ units[/tex]
step 2
Find the value of y
we know that
In the right triangle of the figure
[tex]sin(30^o)=\frac{y}{12\sqrt{3}}[/tex] ----> opposite side divided by hypotenuse (SOH)
Remember that
[tex]sin(30^o)=\frac{1}{2}[/tex]
substitute
[tex]\frac{1}{2}=\frac{y}{12\sqrt{3}}[/tex]
[tex]y=6\sqrt{3}\ units[/tex]
