## Trigonometry (11th Edition) Clone

$45^{o},\quad 225^{o}$
Reference Angle $\theta^{\prime}$ for $\theta$ in $(0^{\mathrm{o}},\ 360^{\mathrm{o}})$ $\left[\begin{array}{lllll} Quad.: & I & II & III & IV\\ \theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta \end{array}\right]$ Tangent is positive in quadrants I and III . Browsing through: Function Values of Special Angles, we find that $\tan 45^{o}=1$, so the reference angle is $45^{o}$ In quadrant I, $\theta=45^{o}$ In quadrant III, $\theta^{\prime}=\theta-180^{o}$ so $\theta=180^{o}+\theta^{\prime}=180^{o}+45^{o}=225^{o}$