Make use of the Law of Sines:
a/sinA=b/sinB=c/sinC
Solving for B
Law of Sines
[1]" "a/sinA=b/sinB
Isolate sinB.
[2]" "cancela/cancelsinA*(cancel((sinA))(sinB))/cancela=b/cancelsinB*((sinA)cancel((sinB)))/a
[3]" "sinB=(bsinA)/a
Plug in the values of a, A, and b.
[4]" "sinB=(54sin(37°))/49
Get the inverse function. Solve using a calculator
[5]" "hArrcolor(blue)(B=arcsin((54sin(37°))/49)~~41.546274420887°)
Solving for C
The sum of all interior angles of a triangle is 180°.
[1]" "A+B+C=180°
Isolate C
[2]" "C=180°-A-B
Plug in the values of A and B.
[3]" "C=180°-37°-arcsin((54sin(37°))/49)
Solve using a calculator.
[4]" "color(blue)(C=143°-arcsin((54sin(37°))/49)~~101.453725579113°)
Solving for c
Law of Sines
[1]" "c/sinC=a/sinA
Isolate c.
[2]" "c/cancelsinC*cancelsinC=a/sinA*sinC
[3]" "c=(asinC)/sinA
Plug in the values of A, a, and C.
[4]" "c=(49sin(143°-arcsin((54sin(37°))/49)))/sin(37°)
Solve using a calculator.
[5]" "color(blue)(c=(49sin(143°-arcsin((54sin(37°))/49)))/sin(37°)~~79.7989130980759269)