## Trigonometry 7th Edition

reference angle = $15.17^o$ Refer to the image below for the drawing.
Convert the angle measurement to degrees by mukltiplying $10'$ by $\dfrac{1^o}{60'}$ to obtain: $195^o + 10' \cdot \dfrac{1^o}{60'}=195^o + 0.1\overline{6^{\circ}}\approx 195.17^o$ The angle is positive so from the positive x-axis, move $195.17$ degrees counterclockwise. (refer to the attached image in the answer part above) The terminal side of the angle is in Quadrant III. The reference angle of an angle $\theta$ in Quadrant II can be found using the formula $\theta-180^o$ Thus, the reference angle of the given angle is: $=195.17^o -180^o \\=15.17^o$