Question

Help with khan academy problem, law of sines

Answer 1

Answer:

104°

Step-by-step explanation:

$\frac{sin~B}{18} =\frac{sin~29}{9} \\sin~B=\frac{18}{9} \times sin~29\\sin~B=2~sin~29\\B=sin^{-1}(2 ~sin~29)\\=104^\circ$