[tex]\textit{Law of Cosines}\\\\ \cfrac{a^2+b^2-c^2}{2ab}=cos(C)\implies cos^{-1}\left(\cfrac{a^2+b^2-c^2}{2ab}\right)=\measuredangle C \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ cos^{-1}\left(\cfrac{90^2+55^2-50^2}{2(90)(55)}\right)=\measuredangle C\implies cos^{-1}\left( \cfrac{8625}{9900} \right)=\measuredangle C \\\\\\ cos^{-1}\left( \cfrac{115}{132} \right)=\measuredangle C\implies 29.4^o\approx \measuredangle C[/tex]
Answer:
C=29.4°
Step-by-step explanation:
[tex]c=50\:ft\\\\a=90\:ft\\\\b=55\:ft[/tex]
Using law of Cosines
[tex]C=cos^{-1}\left(\cfrac{a^2+b^2-c^2}{2ab}\right)\\\\C=cos^{-1}\left(\cfrac{90^2+55^2-50^2}{2*90*55\right)}\\\\C=29.4^{o}[/tex]
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