Answer:
[tex]DF=15, DE=5\sqrt{3}[/tex]
Step-by-step explanation:
[tex]sin(60)=\frac{DF}{EF} =\frac{DF}{10\sqrt{3} }\\DF=10\sqrt{3} sin(60)=10\sqrt{3} \times \frac{\sqrt{3} }{2} =15\\cos(60)=\frac{DE}{EF} =\frac{DE}{10\sqrt{3} }\\DE=10\sqrt{3} cos(60)=10\sqrt{3} \times \frac{1 }{2} =5\sqrt{3}[/tex]
