help geometry

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

Hello Guest!

sine law:

$\frac{sinB}{\overline {CA}}=\frac{sinA}{\overline{CB}}$

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

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

$\angle B=27.575°$

  !