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)#