## Trigonometry (11th Edition) Clone

$54.2^{o},\qquad 234.2^{o}$
reference angles: $\left[\begin{array}{lllll} Quadr.: & I & II & III & IV\\ \theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta \end{array}\right]$ ------------------ The calculator, set to DEGREES, for $\tan^{-1}$ ($1.3865342$) returns 54.2000000983 $\approx 54.2^{o}$ This is the reference angle, in quadrant I. Since tangent is also positive in quadrant III, and for $\theta$ in quadrant III, we calculate the reference angle with $\theta^{\prime}=\theta-180^{o}$, we find the angle $\theta$ with $\theta=180^{\mathrm{o}}+\theta^{\prime}$ $=180^{\mathrm{o}}+54.2^{o}=234.2^{o}$