5. In \( \triangle \mathrm{OPQ}, o=840 \mathrm{~cm}, p=940 \mathrm{~cm} \) and \( q=740 \mathrm{~cm} \). Find the measure of \( \angle \mathrm{O} \) to the nearest degree.

To find the measure of \( \angle O \) in triangle \( \triangle OPQ \), we can use the Law of Cosines, which states: \[ \cos(O) = \frac{p^2 + q^2 - o^2}{2pq} \] Substituting the given values: \[ p = 940\, \text{cm}, \quad q = 740\, \text{cm}, \quad o = 840\, \text{cm} \] First, calculate \( p^2 + q^2 - o^2 \): \[ p^2 = 940^2 = 883600 \] \[ q^2 = 740^2 = 547600 \] \[ o^2 = 840^2 = 705600 \] Now combine these: \[ p^2 + q^2 - o^2 = 883600 + 547600 - 705600 = 725600 \] Next, calculate \( 2pq \): \[ 2pq = 2 \times 940 \times 740 = 1383200 \] Now substitute back into the cosine formula: \[ \cos(O) = \frac{725600}{1383200} \approx 0.524 \] To find \( \angle O \), we take the inverse cosine: \[ O \approx \cos^{-1}(0.524) \approx 58.6^\circ \] Rounding to the nearest degree: \[ \angle O \approx 59^\circ \] Therefore, the measure of \( \angle O \) is approximately \( 59^\circ \).