Question

Check my work? Solving triangle

Answer 1

A = 32 degrees
a = 19
b = 14

Answer 2

\(\sf\dfrac{sin(32)}{19}~=~\dfrac{sin(B)}{14}\)

Answer 3

\(\sf\dfrac{14~sin(32)}{19}~=~sin(b)\)

Answer 4

\(\sf arcsin(\dfrac{14~sin(32)}{19})~=~B\)

Answer 5

B = 23 degrees

Answer 6

Cool.

Answer 7

\(\sf\dfrac{14}{sin(23)}~=~\dfrac {c}{sin(125)}\)

Answer 8

\(\sf\dfrac{14~sin(125)}{sin(23)}~=~c\)

Answer 9

c = 29.4

Answer 10

Seems great to me.
Also another triangle doesn't exist because. B'=180-23=157
and B'+A is over 180 deg
So only the triangle you found exists.

Answer 11

Thanks! :)