Answer:
Answer is given below with explanations.
Step-by-step explanation:
[tex]given \: in \: a \: triangle \: \\ AC = 11 \: ft \: \: and \: angle \: B = 45 \: degrees \\ to \: find \: BC \: we \: use \: sin \: theta \\ we \: know \: sin \: theta = \frac{opposite \: side}{hypotenuse} \\ here \: theta \: = 45 \: degrees \\ sin(45) = \frac{11}{BC} \\ \frac{1}{ \sqrt{2} } = \frac{11}{BC} \\ BC = 11 \sqrt{2} [/tex]
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