Question

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

Answer 1

Use the law of sines, which is
`a/(sin/_A) = b/(sin/_B)`

The small `a` stands for the side opposite to `/_A`, which is `CB`.
The small `b` stands for the side opposite to `/_B`, which is `AC`.

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)`