Answer on Question#39428, Math, Trigonometry
We are given 0<θ<2π0 < \theta < 2\pi0<θ<2π , tgθ=−0.6tg\theta = -0.6tgθ=−0.6 , sinθ<0\sin \theta < 0sinθ<0
Since tgθ=sinθcosθ=−0.6<0tg\theta = \frac{\sin\theta}{\cos\theta} = -0.6 < 0tgθ=cosθsinθ=−0.6<0 and sinθ<0\sin \theta < 0sinθ<0 , thus cosθ>0\cos \theta > 0cosθ>0 , so −π2<θ<π2\frac{-\pi}{2} < \theta < \frac{\pi}{2}2−π<θ<2π . Solution θ=arctan(−0.6)≈−30.96\theta = \arctan(-0.6) \approx -30.96θ=arctan(−0.6)≈−30.96 degrees satisfies given conditions.
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