`A = 120^@, a = 25, b = 24` Use the law of sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answer...

Given: `A=120^@, a=25, b=24`

The Law of Sines:   `a/sin(A)=b/sin(B)=c/sin(C)`

`25/sin(120)=24/sin(B)=c/sin(C)`

`sin(B)=[24sin(120)]/25=.8313`

`B=arcsin(.8313)`

` ` `B=56.24`

`C=180-120-56.24`

`C=3.76`

`25/sin(120)=c/sin(3.76)`

`c=[25sin(3.76)]/sin(120)`

`c=1.89` 

Given: `A=120^@, a=25, b=24`

The Law of Sines:   `a/sin(A)=b/sin(B)=c/sin(C)`

`25/sin(120)=24/sin(B)=c/sin(C)`

`sin(B)=[24sin(120)]/25=.8313`

`B=arcsin(.8313)`

` ` `B=56.24`

`C=180-120-56.24`

`C=3.76`

`25/sin(120)=c/sin(3.76)`

`c=[25sin(3.76)]/sin(120)`

`c=1.89`