Solve the triangle with A 110 degree C 30 degree C 3Solut

Solve the triangle with A = 110 degree, C = 30 degree, C = 3

Solution

2)

A+B+C=180^o

110^o+B+30^o=180^o

=>B=180^o-110^o-30^o

=>B=40^o

by law of sines a_/sinA=b_/sinB=c_/sinC

a_/sin110^o=b_/sin40^o=3_/sin30^o

=>a_/sin110^o=b_/sin40^o=3_/(1/2)

=>a_/sin110^o=b_/sin40^o=6

=>a=6sin110^o,b=6sin40^o

=>a5.638,b3.857

so a5.638,b3.857 ,B=40^o