# How do you find angle B if in triangle ABC, angleA=41^@, AC=23, CB=12?

Dec 17, 2014

Use the law of sines, which is
$\frac{a}{\sin \angle A} = \frac{b}{\sin \angle B}$

The small $a$ stands for the side opposite to $\angle A$, which is $C B$.
The small $b$ stands for the side opposite to $\angle B$, which is $A C$.

Fill this in in the formula, then we get
12/(sin(41^∘)) = 23/(sin/_B)

By using cross-multiplication we find
sin/_B = (23*sin(41^∘))/12

By taking the inverse sine of the whole thing, we get:
/_B =sin^-1( (23*sin(41^∘))/12)