What is the length of AB¯¯¯¯¯, to the nearest tenth of a centimeter? angle A = 42 Angle c = 50 Triangle ABC BC = 12

AB = 13.7 cm.

We will use the law of sines to solve this:
sin A/a = sin B/b = sin C/c

We know that BC, or side a, is 12.  We know that angle A is 42, and angle C (across from AB, or side c) is 50:
(sin 42)/12 = (sin 50)/x

Cross multiply:
x*sin 42 = 12*sin 50

Divide both sides by sin 42:
(x*sin 42)/(sin 42) = (12*sin 50)/(sin 42)
x = 13.74 ≈ 13.7