We will have the following:
First, we have that the problem can be seeing as follows:
Now, knowing this and the the properties of the diagonals we will have that [We use the law of cosines}:
First diagonal [AC} is given by:
[tex]\begin{gathered} AC=\sqrt{16^2+14^2-2cos(41)}\Rightarrow AC=10.67193085... \\ \\ \Rightarrow AC\approx10.7 \end{gathered}[/tex]So, one of the diagonals has an approximate length of 10.7 ft.
Second diagonal [BD] is given by:
[tex]\begin{gathered} BD=\sqrt{16^2+14^2-2(16)(14)cos(139)}\Rightarrow BD=28.10889347... \\ \\ \Rightarrow BD\approx28.1 \end{gathered}[/tex]So, the second diagonal has a length of approximately 28.1 ft.
**Diagonals**
[tex]\begin{gathered} AC\approx10.7ft \\ \\ BD\approx28.1ft \end{gathered}[/tex]
