In $\triangle \mathrm{PQR}, p=99 \mathrm{~cm}, r=97 \mathrm{~cm}$ and $\angle \mathrm{R}=112^{\circ}$. Find all possible values of $\angle \mathrm{P}$, to the nearest degree.

Step 1 ：Use the Law of Cosines to find the length of side q: \(q = \sqrt{p^2 + r^2 - 2pr \cos{R}}\)

Step 2 ：Use the Law of Sines to find the angle P: \(\frac{\sin{P}}{p} = \frac{\sin{R}}{r}\)

Step 3 ：Calculate the values: \(q = \sqrt{99^2 + 97^2 - 2(99)(97) \cos{112^\circ}} \approx 162.50\)

Step 4 ：Calculate the angle P: \(\sin{P} = \frac{99 \sin{112^\circ}}{97} \Rightarrow P \approx 34.39^\circ\)

Step 5 ：\boxed{\text{Final Answer: } P \approx 34^\circ}