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so check the picture below

let's find the missing side first, either b or c, say side "b"

[tex]\bf \textit{Law of Cosines}\\ \quad \\ c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies c = \sqrt{{{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)}\\\\ -----------------------------\\\\ b = \sqrt{{{ a}}^2+{{ c}}^2-(2{{ a}}{{ c}})cos(B)} \\\\\\ b = \sqrt{{{ 79.9}}^2+{{ 46.8}}^2-2(79.9)(46.8)cos(16.3^o)} \\\\\\ b\approx 37.3659169728602838312\implies b\approx 37.37[/tex]

so, b is about 37.37

now, let  us use the law of cosines, to find either angle A or C, hmmm say A

[tex]\bf \textit{Law of Cosines}\\ \quad \\ c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies c = \sqrt{{{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)}\\ \quad \\ \cfrac{{{ a}}^2+{{ b}}^2-c^2}{2{{ a}}{{ b}}}=cos(C)\implies cos^{-1}\left(\cfrac{{{ a}}^2+{{ b}}^2-c^2}{2{{ a}}{{ b}}}\right)=\measuredangle C\\\\ -----------------------------\\\\ [/tex]

[tex]\bf cos^{-1}\left(\cfrac{{{ b}}^2+{{ c}}^2-a^2}{2{{ b}}{{ c}}}\right)=\measuredangle A \\\\\\ cos^{-1}\left(\cfrac{{{ 37.37}}^2+{{ 46.8}}^2-79.9^2}{2(37.37)(46.8)}\right)=\measuredangle A[/tex]

now, what's ∡C? well, all internal angles in a triangle are 180, thus ∡C picks up the slack from A and B, or 180 - A - B

Ver imagen jdoe0001

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