Since it is a right triangle then you can use the Pythagorean theorem to find the measure of the segment DF, like this
[tex]\begin{gathered} a^2+b^2=c^2 \\ \text{ Where a, b are the legs and c is the hypotenuse} \end{gathered}[/tex]
In this case, you have
[tex]\begin{gathered} a=DF \\ b=FE=19 \\ c=DE=28 \end{gathered}[/tex][tex]\begin{gathered} a^2+b^2=c^2 \\ a^2+(19)^2=(28)^2 \\ \text{ Subtract (19)}^2\text{ on both sides of the equation} \\ a^2+(19)^2-(19)^2=(28)^2-(19)^2 \\ a^2=(28)^2-(19)^2 \\ \text{Apply square root on both sides of the equation} \\ \sqrt[]{a^2}=\sqrt[]{(28)^2-(19)^2} \\ a=20.57 \end{gathered}[/tex]
Now to find the measure of angle D, you can use the trigonometric ratio
[tex]\sin (\theta)=\frac{\text{opposite leg }}{\text{hypotenuse}}[/tex]
In this case, you have
[tex]\begin{gathered} \sin (D)=\frac{19}{28} \\ \text{ Apply the inverse function of sin(}\theta\text{) } \\ \sin ^{-1}(\sin (D))=\sin ^{-1}(\frac{19}{28}) \\ D=\sin ^{-1}(\frac{19}{28}) \\ D=42.7\text{ \degree} \end{gathered}[/tex]
Finally, to find the measure of angle E, you can use the trigonometric ratio
[tex]\cos (\theta)=\frac{\text{ adjacent leg}}{hypotenuse}[/tex]
In this case, you have
[tex]\begin{gathered} \cos (E)=\frac{19}{28} \\ \text{ Apply the inverse function of cos(}\theta\text{)} \\ \cos ^{-1}(\cos (E))=\cos ^{-1}(\frac{19}{28}) \\ E=\cos ^{-1}(\frac{19}{28}) \\ E=47.3\text{ \degree} \end{gathered}[/tex]
Therefore,
*The measure of the segment DF is 20.57.
*The measure of angle D is 42.7°.
*The measure of angle E is 47.3.
