Question

Law of Cosines (SSS)In ΔOPQ, o = 700 cm, p = 840 cm and q=620 cm. Find the measure of ∠P to the nearest degree.

Answer 1

79

Explanation:

Answer 2

The Figure above represents the given triangle OPQ with sides o, p, and q

Using Law of Cosines

`p^2=o^2+q^2-2\text{oqCosP}`

Note that o=700cm; p=840cm; q=620cm

Substituting the given values in the expression will give

`840^2=700^2+620^2-2(700*620)Cos\text{ P}`
`\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{ }(168800)/(868000) \\ \text{CosP}=0.1945 \end{gathered}`

To get angle P, we have

`\begin{gathered} P=\cos ^(-1)(0.1945) \\ P=78.78^0 \\ P=79^{0\text{ }}(to\text{ the nearest degree)} \end{gathered}`

Hence∠P to the nearest degree is 79°