In triangle ABC, CA=4sqrt(2), CB=4sqrt(3), and A= \(45\) degrees. What is B in degrees?
+9674 Applying the Law of Sines yields:
\(\dfrac{CA}{\sin B} = \dfrac{CB}{\sin A}\)
After some simplification,
\(\sin B = \dfrac{CA \cdot \sin A}{CB} = \dfrac{4 \sqrt 2 \sin 45^\circ}{4\sqrt 3}\)
Now you can find the value of sin B, and hence the value of B using inverse trigonometric function.
