How do you find the angle $α$ $alpha$ such that the angle lies in quadrant III and $tan α = 8.000$ $tanalpha=8.000$ ?

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How do you find the angle $α$ $alpha$ such that the angle lies in quadrant III and $tan α = 8.000$ $tanalpha=8.000$ ?

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Answer 1 · 2016-07-15T09:30:34

The principal value $α_{P}$ $alpha_P$ , for $tan α > 0$ $tan alpha> 0$ is in the first quadrant. The general value $= n π + α_{P}, n = 0, \pm 1, \pm 2, \pm 3,$ $= npi + alpha_P, n = 0, +-1, +-2, +-3,$ ..Here, the third quadrant (n=1) value = $π + α_{P} . = 270 -^{o}$ $pi+alpha_P.= 270-^o$

Explanation:

Interestingly, 3-sd (rounded) principal value $α_{P}$ $alpha_P$ , for $tan α = 8000$ $tan alpha=8000$ , is ${90.0}^{o}$ $90.0^o$ .

So, the answer is little short of $270^{o}$ $270^o$ .

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How do you find the angle $α$ $alpha$ such that the angle lies in quadrant III and $tan α = 8.000$ $tanalpha=8.000$ ?

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How do you find the angle alphaαalpha such that the angle lies in quadrant III and tanalpha=8.000tanα=8.000tanalpha=8.000?

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How do you find the angle $α$ $alpha$ such that the angle lies in quadrant III and $tan α = 8.000$ $tanalpha=8.000$ ?