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

Dec 17, 2014

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:
$\frac{a}{\sin \angle A} = \frac{b}{\sin \angle B} = \frac{c}{\sin \angle C}$
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)