The Figure above represents the given triangle OPQ with sides o, p, and q
Using Law of Cosines
[tex]p^2=o^2+q^2-2\text{oqCosP}[/tex]Note that o=700cm; p=840cm; q=620cm
Substituting the given values in the expression will give
[tex]840^2=700^2+620^2-2(700\times620)Cos\text{ P}[/tex][tex]\begin{gathered} 705600=490000+384400-868000\text{CosP} \\ 705600=874400-868000\text{CosP} \\ 868000\text{CosP}=874400-705600 \\ 868000\text{CosP}=168800 \\ \text{CosP}=\text{ }\frac{168800}{868000} \\ \text{CosP}=0.1945 \end{gathered}[/tex]To get angle P, we have
[tex]\begin{gathered} P=\cos ^{-1}(0.1945) \\ P=78.78^0 \\ P=79^{0\text{ }}(to\text{ the nearest degree)} \end{gathered}[/tex]Hence∠P to the nearest degree is 79°
