Answer: [tex]x=18[/tex] ; [tex]y=6\sqrt{3}[/tex]
Step-by-step explanation:
You can write two expressions that are in terms of 'x' and 'y'. This is a right triangle, so Pythagorean's theorem can be used. We can also use the angle to find the sine of 30 degrees.
Pythagorean's Theorem:
The values 'a' and 'b' are the two shorter sides of the triangle, while the value 'c' is the longest side of the triangle; the hypotenuse.
[tex]a^2+b^2=c^2[/tex]
[tex]x^2+y^2=(12\sqrt{3} )^2[/tex]
[tex]x^2+y^2=12^2*\sqrt{3} ^2=432[/tex]
Sine of the Angle:
Sine is defined to be the opposite side divided by the hypotenuse. Let's take the sine of 30 degrees:
[tex]sin(\alpha )=opposite/hypotenuse[/tex]
[tex]sin(30)=y/12\sqrt{3}[/tex]
[tex]\frac{1}{2} =y/12\sqrt{3}[/tex]
[tex]y=6\sqrt{3}[/tex]
Plug this value of 'y' into the first expression derived from Pythagorean's theorem:
[tex]x^2+y^2=432[/tex]
[tex]x^2+(6\sqrt{3} )^2=432[/tex]
[tex]x^2+108=432[/tex]
[tex]x^2=432-108=324[/tex]
[tex]x=\sqrt{324}[/tex]
[tex]x=18[/tex]
Use this value of 'x' to solve for 'y' in the expression from Pythagorean's theorem:
[tex](\sqrt{324})^2+y^2=432[/tex]
[tex]324+y^2=432[/tex]
[tex]y^2=432-324=108[/tex]
[tex]y=\sqrt{108}[/tex]
[tex]y=\sqrt{36*3}[/tex]
[tex]y=6\sqrt{3}[/tex]
