In $\triangle \mathrm{OPQ}, q=18 \mathrm{~cm}, o=96 \mathrm{~cm}$ and $\angle \mathrm{P}=142^{\circ}$. Find the length of $p$, to the nearest centimeter.
Answer
\(\boxed{p = 111\text{ cm}}\)
Step 1 ：Given: \(o = 96\), \(q = 18\), and \(\angle P = 142^\circ\)
Step 2 ：Convert \(\angle P\) to radians: \(\angle P_{rad} = \frac{142 \times \pi}{180} = 2.4784\)
Step 3 ：Use the Law of Cosines to find the length of side p: \(p^2 = o^2 + q^2 - 2 \times o \times q \times \cos(P)\)
Step 4 ：Plug in the given values: \(p^2 = 96^2 + 18^2 - 2 \times 96 \times 18 \times \cos(2.4784)\)
Step 5 ：Calculate \(p^2\): \(p^2 = 12263.3652\)
Step 6 ：Find the length of side p: \(p = \sqrt{12263.3652} = 110.7401\)
Step 7 ：Round p to the nearest centimeter: \(p \approx 111\)
Step 8 ：\(\boxed{p = 111\text{ cm}}\)
