Question

How do you find `angleB` if in triangle ABC, `a = 15`, `b = 20` , and `angle A=30^@`?

Answer 1

The answer is: `/_B = sin^-1((20*sin(30^∘))/15)`

You have to use the law of sines, just as you thought. This law is:
`a/(sin/_A) = b/(sin/_B) = c/(sin/_C)`
We only need the first equation here, let's fill in the numbers.
`15/sin(30^∘) =20/(sin/_B)`
We use cross-multiplication, then we find:
`sin/_B = (20*sin(30^∘))/15`
then by taking the inverse sine, we find
`/_B = sin^-1((20*sin(30^∘))/15)`