Answer:
[tex]let \: \alpha \: be \: the \: 45 \\ \tan( \alpha ) = \frac{p}{b} \\ \tan(45) = \frac{x}{3 \sqrt{2} } \\ 1 = \frac{x}{3 \sqrt{2} } \\ x = 3 \sqrt{2} \\ by \: using \: pythagoras \: therom \\ {y}^{2} = {3 \sqrt{2} }^{2} + {3 \sqrt{2} }^{2} \\ {y}^{2} = 18 + 18 \\ y = \sqrt{36} \\ y = 6[/tex]
Step-by-step explanation:
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