Trisecting an angle

Drop an altitude from D to OB. Label this point E. Because ACE is a right angle and ACD is $60^\circ$ , DCE must be $30^\circ$ . Because all figures are proportional, we can let the sidelength of the equilateral triangle be 2 and because of the $30^\circ, 60^\circ ,90^\circ$ triangle, $CE=\sqrt{3}$ and $DE=1$ . Let $OC=x$ , and $angle COA = 3\theta$ . Assume that $DO$ trisects the angle.

Now we can see that $\tan \theta= \frac {1}{\sqrt{3}+x}$ and $\tan 3\theta=\frac {2}{x}$ . We simplify $\tan3\theta$ into $\frac {3\tan\theta-{\tan\theta}^3}{1-3{\tan\theta}^{2}}$ [by splitting it up into a tangent sum with $2\theta$ and $\theta$ , and splitting the $2\theta$ into a double angle tangent identity]. Because we know what $\tan\theta$ is, we can substitute into the the equation for $\tan3\theta$ and it is an equation with only x as a variable. After solving the equation, you get that $x=\pm 2$ , which leaves that $\tan3\theta = 2/(+/-2)=+/1$ . Since the angle must be between $0^\circ$ and $90^\circ$ , the only viable choice is when it is 1 and that occurs when the angle is $45^\circ$ .

Construct a perpendicular $DE$ on $OB$. Let $AC=CD=AD=a$. Since $\triangle ACD$ is an equilateral triangle, $\angle ACD = 60^{\circ}$ and thus $\angle DCE = 30^{\circ}$. So in right angled $\triangle DCE, DE=a/2$ and $CE=a\sqrt{3}/2$.\ \indent Now by assumption $\angle DOE = \theta /3$, so $OD=\frac{a}{2\sin({\theta /3})}$ and $OE=ODcos(\theta /3)=\frac{a}{2}cot(\theta /3)$. \ \indent Now $OC=OE-CE=\frac{a}{2}(cot(\theta /3) - \sqrt{3})$. Hence $tan(\angle AOC)=tan(\theta)=AC/OC=\frac{2}{(cot(\theta /3) - \sqrt{3})}$.\ \indent Let $f(\theta)=tan(\theta)-\frac{2}{(cot(\theta /3) - \sqrt{3})}$. The function $f$ has only one zero in the range $0$ to $\pi /2$ at $\theta = \pi /4$. Therefore $OD$ is an angle trisector only for $\theta = \pi /4$.

Reflect the point $D$ across $OB$ to a point $E$ . Since $\angle DCA = 60^\circ$ , it follows that $150^\circ = \angle DCO = \angle ECO$ , so $\angle DCE = 60^\circ$ . By the reflection, $CD=CE$ , so $CDE$ is an equilateral triangle.

Consider the rotation of triangle $DOE$ about $O$ by $\angle DOE$ . $E$ gets mapped to $D$ , and $D$ gets mapped to $F$ , which lies on $OA$ . Since $DOF$ is an isosceles triangle, $\angle DFO = \frac {1}{2} (180^\circ - \angle DOF) = 90^\circ - \angle DOC = 90^\circ - \frac {1}{3} \angle COA$ .

By construction, $DA=DC=DE=FD$ , so $DFA$ is an isosceles triangle. Let us determine where $F$ lies. $DOF$ is an isosceles triangle. Observe that $\angle DCO$ is obtuse, so $OD^2 > OC^2 + CD^2 = OC^2 + CA^2 = OA^2$ , hence $F$ is further out from $O$ than $A$ (and also not equal to $A$ ). So, we have $\begin{aligned} \angle DFO = \angle DFA = \angle DAF &= 180^ \circ - 60^ \circ - \angle CAO \\ &= 120^\circ - (90^\circ -\angle COA) \\ &= 30^\circ + \angle COA \\ \end{aligned}$

Equating these two, we get that $90^\circ - \frac {1}{3} \angle COA = 30^\circ + \angle COA$ , or that $\angle COA = 45^\circ$ . Hence, there is only 1 possible value for $\theta$ , and so the sum is $45$ .