Question

ABC is a triangle . AC= 19 cm, BC= 17 cm and triangle BAC = 60 drgees

Answer 1

AC = 19cm
BC = 17cm
angle A= 60

Using sin formula

$\frac{BC}{sin\ A} = \frac{AC}{sin\ B} = \frac{AB}{sin\ C} \\ \\ \frac{17}{sin\ 60} = \frac{19}{sin\ B} \\ \\ sin\ B=\frac{19}{17} *sin\ 60 \\ \\ sin\ B= \frac{19}{17} * \frac{ \sqrt{3} }{2} =0.9679 \\ \\ B=sin^{-1}(0.9679)=75.48$

So angle ABC or angle B is 75.48 degree.

Answer 2

Use the law of sines to find angle B
angle you want /ac=angle you know (a)/bc
All you need is a/sin(A)=b/ sin(B)
Solve in this way. You will get the answer as 75.48 degree