Answer: [tex]DE\approx6.0[/tex]
Step-by-step explanation:
Observe in the figure given in the exercise that four right triangles are formed.
In this case you can use the following Trigonometric Identity to solve this exercise:
[tex]cos\alpha=\frac{adjacent}{hypotenuse}[/tex]
From the figure you can identify that:
[tex]\alpha =53\°\\\\hypotenuse=10\\\\adjacent=BE=DE[/tex]
Then, you can substitute values:
[tex]cos(53\°)=\frac{DE}{10}[/tex]
The next step is to solve for DE in order to find its value. This is:
[tex]10*cos(53\°)=DE\\\\DE=6.01[/tex]
Finally, rounding the result to the nearest tenth, you get that this is:
[tex]DE\approx6.0[/tex]
