use these formulea
a^2=b^2+c^2-(2bcCosA)
then to get angle C
sinA/a=sinC/a...(1)
to get angle B
sinB/b=sinC/c....(2)
there you go a big hint
Here it just says this:
Solve triangle ABC. (If an answer does not exist, enter DNE. Round your answers to one decimal place.)
b = 67, c = 33, ∠A = 70°
2 answers
or, once you have A and C, B = 180-(A+C)