Solve the triangle ABC if b 125 c 165 403 degree a Assum

Solve the triangle ABC if b = 125, c = 165, = 403 degree. a = Assume A is opposite side a, B is opposite side b, and C is opposite side c.

given b =125,c=165,B=40^o

by law of sines

b/sinB =c/sinc

125/sin(40^o) =165/sinC

sinC=(165/125)sin(40^o)

sinC=0.84848

C=58^o,122^o

when B=40,C=58

A+B+C =180

A=180-40-58

A=82^o

a/sinA =b/sinB

a/sin82^o =125/sin40^o

a =125(sin82^o/sin40^o)

a=192.6

when B=40,C=122

A+B+C =180

A=180-40-122

A=18^o

a/sinA =b/sinB

a/sin18^o =125/sin40^o

a =125(sin18^o/sin40^o)

a=60.1

2 sets of solutions are

a=192.6 ,A=82^o,C=58^o

a=60.1 ,A=18^o,C=122^o