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^@ ,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#