trig
posted by Matt .
Solve sin θ = 0.204 for 90º < θ < 270º. Give your answer to the nearest tenth of a degree.
Honestly have no idea how to do this.

so Ø must be in quadrant II or III , but in II the sine would be positive,
so Ø is in III
get the "angle in standard position" by using +.204
sin^1 (.204) = 11.77°
so in III, the angle is 180 + 11.77
= 191.77°
= 191.8° correct to one decimal
check with your calculator:
sin 191.77 = .20398.. , ok then !