How do you solve the right triangle given b=1, c=2, and A= 120?

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Answer 1 · 2016-06-26T15:38:54

$a = 1.4, B = {37.76}^{\circ}, C = {22.24}^{\circ}$ $a=1.4,B=37.76^@ ,C=22.24^@$

By properties of triangle we know,
In $DeltaABC ,"a,b,c are opposite sides of angles A,B,C"$

Given A=120, C=180-120-B=60-B

$a/sinA=b/sinB=c/sinC$

$a/sin120=1/sinB=2/sin(60-B)$
So
$=>sin(60-B)/sinB=2$

$=>(sin60cosB)/sinB-cos60=2$

$=>sqrt3/2cotB-1/2=2$

$=>sqrt3cotB-1=4$

$=>cotB=5/sqrt3$

$=>tanB=sqrt3/5$

$=>B=tan^-1(sqrt3/5)=37.76^@$

$:.C=60-37.76=22.24^@$

Now

$a=sin120/sinB=sqrt3/(2*sin(37.76))=1.4$