# trig

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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.

• trig -

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

sin 191.77 = -.20398.. , ok then !

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