​find angle c with trig

There are several approaches, but here's mine

Let's find  AD  using the Law of Sines

AD / sin ABD  = AB / sin BDA

AD /  sin(70)  =  75 / sin (90)

AD  =  75 sin (70)  

So....DC  =   90  - 75sin (70)  

Similarly.... we can find BC  as  75 sin (20)   

So......using the tangent

Tan ( C )   =    BC / DC

Tan (C)  =   [ (75) sin (20) /  (   90  - 75sin (70) ) ]

Using the arctangent

arctan   [ 75 sin (20) /  (   90  - 75sin (70) ) ]  = C  ≈  .92 rads  ≈  52.7°

find angle c with trig

\(\begin{array}{|rcll|} \hline \tan(B) &=& \dfrac{c \cdot \sin(A)}{b-c\cdot \cos(A)} \quad & | \quad c = 75\ cm \quad b = 90\ cm \quad A = 20^{\circ} \\\\ \tan(B) &=& \dfrac{75 \cdot \sin(20^{\circ})}{90-75\cdot \cos(20^{\circ})} \\\\ \tan(B) &=& \dfrac{75 \cdot 0.34202014333}{90-75\cdot 0.93969262079} \\\\ \tan(B) &=& 1.31390875033 \\\\ B &=& \arctan(1.31390875033) \\\\ \mathbf{B} & \mathbf{=} & \mathbf{52.7256774464^{\circ}} \\ \hline \end{array}\)