Any math geniuses out there????? Trigonometry question!

I dont know what to do!! Please help.

Solve triangle ABC assuming the given measures. (notation is standard that means a,b and c denote the sides opposite angles A,B and C, respectively)

Exercise

a=20; b=40; A=30º

Anything at all will help. Thankss!

use the sine law
a/sinA=b/sinB=c/sinC

20/sin(30)=40/sin(B)
sin(B)=1
B=90
so 90+30=120
180-120=60
A=30
B=90
C=60

edit:just realised you need the length of sides
so use the sine law again for c
a/sinA=c/sinC

20/sin(30)=c/sin(60)
c=8.66
or sqrt(75)

a/sinA = b/sinB

so 20/sin30 = 40/sinB
this becomes arcsin (40sin30/20)=B
so B = 90º

so C = 60º

so c = sqrt(20^2+40^2) = 20(sqrt5)

so area = 0.5*20*40 = 800

Law of Sines : a / sin A = b / sin B = c / sin C

Law of Cosines :

a² = b² + c² − 2bc cos A

b² = a² + c² − 2ac cos B

c² = a² + b² − 2ab cos C

sin30 = 0.5.

do substitutions repeatedly, to get other sides and angles.

is the triangle is right-angled triangle?