The given triangle is DEF where angle E is a right angle.
The value of DE is 2.1 and EF is 5.3.
As the triangle is a right angle triangle, hence applying the pythagoras theorem,
![\begin{gathered} DF^2=DE^2+EF^2 \\ DF^2=(2.1)^2+(5.3)^2 \\ DF^2=4.41+28.09 \\ DF^2=32.5 \\ DF=\sqrt[]{32.5} \\ DF=5.70 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/1sr1abzylka25p6pbdfn7fqzyixtwo8bgt.png)
Thus, the value of the side DF is 5.70 units.
The value of angle D and F can be determined as,
![\begin{gathered} \tan D=(EF)/(DE) \\ \tan D=(5.3)/(2.1) \\ \angle^{}D=68.38^(\circ) \\ \angle D\approx68^(\circ) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ignghy74xq4g0f2vzep8t6pixv7nmjtkd9.png)
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