How do you solve $cot θ = - 0.23$ ?

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Answer 1 · 2018-04-08T05:35:35

$theta = (n pi - 1.345)^c " for " n in ZZ$

$cot theta = -0.23$

$tan theta = 1 / cot theta = -1/0.23$

$theta = tan^-1 ((-1/0.23) = -1.345^c$

$theta = (-1.345 * 180) / pi = -77.05^@$

$"Tan is negative in II & IV quadrants"$

$:. theta = 180 - 77.05 = 102.95^@, 360 - 77.95 = 282.95^@$

Generalizing,

$theta = (n pi - 1.345)^c " for " n in ZZ$

How do you solve cot θ = - 0.23cot θ = - 0.23?