Answer:
![x=8, y=8\sqrt{3}](https://tex.z-dn.net/?f=x%3D8%2C%20y%3D8%5Csqrt%7B3%7D)
Step-by-step explanation:
To find y, we need to use the Law of Sines
Recall that the Law of Sines states: ![\frac{a}{sin(A)} =\frac{b}{sin(B)}](https://tex.z-dn.net/?f=%5Cfrac%7Ba%7D%7Bsin%28A%29%7D%20%3D%5Cfrac%7Bb%7D%7Bsin%28B%29%7D)
In this case, 16 is a, 90 is A, 60 is B, and y is b.
Knowing this, we can substitute in our known values, solve for y, and simplify.
![\frac{16}{sin(90)} =\frac{y}{sin(60)} \\\\y=\frac{16sin(60)}{sin(90)} \\\\y=\frac{16*\frac{\sqrt{3} }{2} }{1} ={16*\frac{\sqrt{3} }{2}=8\sqrt{2}](https://tex.z-dn.net/?f=%5Cfrac%7B16%7D%7Bsin%2890%29%7D%20%3D%5Cfrac%7By%7D%7Bsin%2860%29%7D%20%5C%5C%5C%5Cy%3D%5Cfrac%7B16sin%2860%29%7D%7Bsin%2890%29%7D%20%5C%5C%5C%5Cy%3D%5Cfrac%7B16%2A%5Cfrac%7B%5Csqrt%7B3%7D%20%7D%7B2%7D%20%7D%7B1%7D%20%3D%7B16%2A%5Cfrac%7B%5Csqrt%7B3%7D%20%7D%7B2%7D%3D8%5Csqrt%7B2%7D)
Now that we know the value of two of the sides, we can use the Pythagorean Theorem to solve for x.
For simplicity, I will put the 8 back into the radical to make things easier.
![y=8\sqrt{3}=\sqrt{192}](https://tex.z-dn.net/?f=y%3D8%5Csqrt%7B3%7D%3D%5Csqrt%7B192%7D)
Recall that the Pythagorean Theorem states: ![a^2+b^2=c^2](https://tex.z-dn.net/?f=a%5E2%2Bb%5E2%3Dc%5E2)
In this case, x is a, y is b, and 16 is c.
Now we can sub in these values
![x^2+(\sqrt{192})^{2} =16^2](https://tex.z-dn.net/?f=x%5E2%2B%28%5Csqrt%7B192%7D%29%5E%7B2%7D%20%3D16%5E2)
Now we can solve for x and simplify
![x^2+(\sqrt{192})^{2} =16^2\\\\x^2=16^2-(\sqrt{192})^{2}\\\\x=\sqrt{16^2-(\sqrt{192})^{2}} \\\\x=\sqrt{256-192} \\\\x=\sqrt{64} \\\\x=8](https://tex.z-dn.net/?f=x%5E2%2B%28%5Csqrt%7B192%7D%29%5E%7B2%7D%20%3D16%5E2%5C%5C%5C%5Cx%5E2%3D16%5E2-%28%5Csqrt%7B192%7D%29%5E%7B2%7D%5C%5C%5C%5Cx%3D%5Csqrt%7B16%5E2-%28%5Csqrt%7B192%7D%29%5E%7B2%7D%7D%20%5C%5C%5C%5Cx%3D%5Csqrt%7B256-192%7D%20%5C%5C%5C%5Cx%3D%5Csqrt%7B64%7D%20%5C%5C%5C%5Cx%3D8)