Solve for a in the oblique triangle

Question

Using the Law of Sines, ABC: AB=30:AC=15, and the angle B=20 degrees.

Answer 1

I'm guessing angle "b" is between "ab" and "ac".

To label this triangle correctly: side "c"=30; side "b"=15; angle "α"=20°.

Side "a" shortest; "b" mid; "c" hypotenuse.

Angles: "α" opposite 'a'; "β" opposite "b"; "γ" opposite "c".

To find side "a": (b^2+c^2-2bc(cosα))^.5=a.

So: (15^2+30^2-(2*15*30)cos20)^.5=a

(225+900 -900 * .939)^.5=16.71="a"

To get an angle use: a/sinα=b/sinβ=c/sinγ.

16.71/.342=48.86=15/sinβ;

" =.307; β=17.878..........

"γ"=142.122..........

Answer 2

Sin B/15 = sin C/30

sin 20/15 = sin C /30

30*sin20/15 = sin C

C = 43.16

A = 180 - 20 - 43.16 = 116.84

sin A/a = sin B/15

sin 116.84/a = sin 20/15

a = 15 * sin 116.84/sin 20 = 39.13