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âg uls puy s ZE =gQuel+Qo93s JI |
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Answer» secΘ +tanΘ = 2/3 1/cosΘ + sinΘ/cosΘ = 2/3 sinΘ + 1 = 2/3 cosΘ sinΘ = 2/3 cosΘ – 1 Θ = -0.39479 ± 1.176n Or Θ = -22.6198 degrees, which is in Quadrant 4 |
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