Question

Use trigonometric ratios to solve the right triangle.

Answer 1

Answer:

$DF=15, DE=5\sqrt{3}$

Step-by-step explanation:

$sin(60)=\frac{DF}{EF} =\frac{DF}{10\sqrt{3} }\\DF=10\sqrt{3} sin(60)=10\sqrt{3} \times \frac{\sqrt{3} }{2} =15\\cos(60)=\frac{DE}{EF} =\frac{DE}{10\sqrt{3} }\\DE=10\sqrt{3} cos(60)=10\sqrt{3} \times \frac{1 }{2} =5\sqrt{3}$