In $\triangle \mathrm{DEF}, d=6.9 \mathrm{~cm}, f=5 \mathrm{~cm}$ and $\angle \mathrm{F}=44^{\circ}$. Find all possible values of $\angle \mathrm{D}$, to the nearest 1oth of a degree.

Step 1 ：Use the Law of Sines: \(\frac{d}{\sin(D)} = \frac{f}{\sin(F)}\)

Step 2 ：Plug in the given values: \(\sin(D) = \frac{6.9 \times \sin(44)}{5}\)

Step 3 ：Calculate the sine of angle D: \(\sin(D) \approx 0.9586\)

Step 4 ：Find angle D using arcsine: \(D \approx 74.0°\) and \(D \approx 106.0°\)

Step 5 ：\(\boxed{\text{Final Answer: Angle D has two possible values: approximately 74.0° and 106.0°}}\)