Find the missing lengths please help for a grade

Answer:
The answer is
[tex]x = 10 \sqrt{3} \\ y = 5 \sqrt{3} [/tex]
Step-by-step explanation:
[tex] \sin(60) = \frac{15}{x} \\ x= \frac{15}{ \sin(60) } = 10 \sqrt{3} \\ \sin(30) = \frac{y}{10 \sqrt{3} } \\ y = 10 \sqrt{3} \sin(30) \\ y = 10 \sqrt{3} \times .5 \\ y = 5 \sqrt{3} [/tex]