Question

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Answer 1

Aplicăm teoremă lui Pitagora

${BC}^{2} = {AC}^{2} + {AB}^{2}$

${BC}^{2} = {(2 \sqrt{2}) }^{2} + {2}^{2}$

${BC}^{2} =8 + 4$

${BC}^{2} = 12$

${BC}^{} = \sqrt{12} = 2 \sqrt{3}$

$\cos {C}^{} = \frac{AC}{BC} = \frac{2 \sqrt{2} }{2 \sqrt{3} } = \frac{ \sqrt{2} }{ \sqrt{3} } = \frac{ \sqrt{6} }{3}$