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Answer:
Option A
Step-by-step explanation:
Given:
[tex] \sf \therefore \: cos \: 45 \degree = \frac{base}{hypotenuse} = \frac{9}{x} [/tex]
[tex] \sf \cos(45 \degree) = \frac{1}{ \sqrt{2} } [/tex]
[tex] \sf \frac{1}{ \sqrt{2} } = \frac{9}{x} [/tex]
[tex] \therefore \: \rm \: x = 9 \sqrt{2} [/tex]