help geometry

In  triangle ABC, CA=4sqrt(2), CB=4sqrt(7), and A=60 degrees. What is B in degrees?

Hello Guest!

sine law:

\(\frac{sinB}{\overline {CA}}=\frac{sinA}{\overline{CB}}\)

\(\frac{sinB}{4\sqrt{2} }=\frac{sin60°}{4\sqrt{7}}\\ \frac{sinB}{4\sqrt{2} }=\frac{\frac{1}{2}\sqrt{3} }{4\sqrt{7}}\\\)

\(sinB=\dfrac{\frac{1}{2}\sqrt{3}\cdot4\sqrt{2}}{4\sqrt{7}}=\dfrac{1}{2}\sqrt{\frac{6}{7}}\\ \angle B=asin\ (\dfrac{1}{2}\sqrt{\frac{6}{7}})\)

\(\angle B=27.575°\)

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