1. **Problem statement:** Given three lines AB, EF, and CD intersecting at point O, with AB perpendicular to CD, and the ratio of angles $m\angle 1 : m\angle 2 = 1 : 2$, find $m\angle EOB$. 2. **Known facts and formulas:** - Since AB is perpendicular to CD, $m\angle AOC = 90^\circ$. - Angles 1 and 2 are adjacent to these lines and satisfy $m\angle 1 : m\angle 2 = 1 : 2$. - The sum of angles around point O on a straight line is $180^\circ$. 3. **Assign variables:** Let $m\angle 1 = x$, then $m\angle 2 = 2x$. 4. **Analyze the angles:** - Since AB is vertical and CD is horizontal, $\angle AOC = 90^\circ$. - Angles 1 and 2 are parts of this right angle split by EF. - Therefore, $x + 2x = 3x = 90^\circ$. 5. **Calculate $x$:** $$3x = 90^\circ$$ $$x = \frac{90^\circ}{3} = 30^\circ$$ 6. **Find $m\angle EOB$:** - $\angle EOB$ is the angle between OE and OB. - Since $m\angle 1 = 30^\circ$ and $m\angle 2 = 60^\circ$, and considering the geometry, $m\angle EOB = 180^\circ - m\angle 2 = 180^\circ - 60^\circ = 120^\circ$. **Final answer:** $$m\angle EOB = 120^\circ$$