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

Question

4463 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-12-17T21:30:43

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:
$/_B =sin^-1( (23*sin(41^∘))/12)$

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