Reflection across a circular arc

Let $O(0,0)$ be the origin of the $xy$ -plan, $A(5,0)$ , and $B(7\cos 70^\circ, 7\sin 70^\circ)$ . Let the angle to be found $\angle COA = \theta$ . Then $C (2\cos \theta, 2 \sin \theta)$ . From

$\begin{aligned} \angle ACD & = \angle BCD \\ \angle COA + \angle CAO & = \angle BCE - \angle DCE & \small \blue{\text{where }CE||OA} \\ \theta + \tan^{-1} \left(\frac {2\sin \theta}{5-2\cos \theta}\right) & = \tan^{-1} \left(\frac {7\sin 70^\circ - 2\sin \theta}{7\cos 70^\circ -2\cos \theta}\right) - \theta & \small \blue{\text{Note that }\angle DCE=\angle COA = \theta} \end{aligned}$

Solving the above numerically, we have $\theta \approx \boxed{32.3867}$ .

Applying Sine Rule to the $\triangle 's$ $AOC$ and $BOD$ we get $4900sin^4α-1315.5697sin^3α-4508.2343sin^2α+342.8189sinα+908.6299=0$ where $α$ is the required angle. From this we get the value of $sinα$ in the given region to be equal to $\boxed {32.3846}$

Constants:

$\theta_{EOF} = 70^\circ \\ R = 2 \\ A_x = 5 \\ A_y = 0 \\ B_x = 7 \, \cos \theta_{EOF} \\ B_y = 7 \, \sin \theta_{EOF} \\ \theta = \angle C O A$

Coordinates of point $C$ :

$C_x = R \, \cos \theta \\ C_y = R \, \sin \theta$

Incident vector from $A$ to $C$ :

$v_x = C_x - A_x \\ v_y = C_y - A_y$

Normal and tangent vectors to circle:

$N_x = \cos \theta \\ N_y = \sin \theta \\ T_x = -N_y \\ T_y = N_x$

Express $\vec{v}$ in terms of the normal and tangent vectors:

$v_x = \alpha \, N_x + \beta \, T_x \\ v_y = \alpha \, N_y + \beta \, T_y$

Solve the system for $(\alpha,\beta)$ and construct the reflected vector $\vec{v}_2$ :

$v_{2x} = -\alpha \, N_x + \beta \, T_x \\ v_{2y} = -\alpha \, N_y + \beta \, T_y$

Equation of line $OB$ :

$y = m_{OB} \, x$

Trace the reflected ray from point $C$ until it intersects line $OB$ . Determine the travel distance until intersection.

$C_y + \gamma \, v_{2y} = m_{OB} (C_x + \gamma \, v_{2x}) \\ \gamma = \frac{m_{OB} C_x - C_y}{v_{2y} - m_{OB} v_{2x}}$

Intersection point:

$x_i = C_x + \gamma \, v_{2x} \\ y_i = C_y + \gamma \, v_{2y}$

Sweep the parameter $\theta$ and store the value of $\theta$ for which the following is true.

$x_i = B_x \\ y_i = B_y$

This results in $\theta \approx 32.38^\circ$

$Let ~\angle COA=X.\\ \angle ACD = \angle BCD,~~so~ \angle ACO = \angle BCO= say~ Y.\\ ~~~\\ Applying~Cos~Rule~to~\Delta~s~~AOC~~and~~BOC, \\ AC^2=25+4-2*5*2*CosX..........(1)\\ BC^2=49+4-2*7*2*Cos(70-X)......(2)\\ ~~~~\\ Applying~Sin~Rule~to~\Delta~s~~AOC~~and~~BOC, \\ AC=5* SinX/SinY ..........(3) \\ BC=7*Sin(70-X)/ SinY ..........(4) \\ ~~~\\ Subtracting~SinY~~of~(2)~and~(4) ~~from~~ SinY~~of~(1)~and~(3),\\ (5*SinX)^2/(29-2*5*2*CosX) - (7*Sin(70-X)^2/(51-2*7*2*Cos(70-X))=0 \\ ~~\\ Solving~through~ Keisan ~Online~ Calculator ~ X=20~to~60, ~and ~refining, \\ we~get~\angle COA=\Large~\color{#D61F06}{32.38672}.$

Let $\phi = \angle EOF = 70^{\circ}$ , and let $\theta = \angle COA$ . Also, let $\Psi = \angle OCA = \angle OCB$ (because $\angle ACD = \angle BCD$ ). Further, let $a = OA , b = OB, r = OC$

Using the Law of Sines in $\triangle AOC$ , we obtain,

$\dfrac{\sin \Psi}{a} = \dfrac{ \sin( \pi - \theta - \Psi ) }{ r } =\dfrac{ \sin( \Psi +\theta ) }{ r }$

Similarly, using the Law of Sines in $\triangle BOC$ , we obtain,

$\dfrac{\sin \Psi}{b} = \dfrac{ \sin( \Psi + \phi - \theta ) }{ r }$

After some simple manipulation, the above two equations can be solved for $\tan \Psi$ , and from the first equation we obtain,

$\tan \Psi = \dfrac{ a \sin \theta }{ r - a \cos \theta }$

Similarly, the second equation yields,

$\tan \Psi = \dfrac{ b \sin(\phi - \theta) }{ r - b \cos(\phi - \theta) }$

Equating the two expressions and cross multiplying, gives,

$a \sin \theta (r - b \cos(\phi - \theta) ) = b \sin(\phi - \theta) ( r - a \cos \theta )$

Expanding $\cos( \phi - \theta )$ and $\sin ( \phi - \theta )$ , the above equation becomes,

$a \sin \theta (r - b \cos \phi \cos \theta - b \sin \phi \sin \theta ) = b (\sin \phi \cos \theta - \cos \phi \sin \theta ) ( r - a \cos \theta )$

Muliplying the expressions out yields,

$a r \sin \theta - a b \cos \phi \sin \theta \cos \theta - a b \sin \phi \sin^2 \theta = b r \sin \phi \cos \theta - b r \cos \phi \sin \theta - a b \sin \phi \cos^2 \theta + a b \cos \phi \sin \theta \cos \theta$

Collecting terms, and using the identities $\cos 2 \theta = \cos^2 \theta - \sin^2 \theta$ and $\sin 2 \theta = 2 \sin \theta \cos \theta$ , we arrive at,

$(b r \sin \phi ) \cos \theta + (- a r - b r \cos \phi) \sin \theta + (-a b \sin \phi )\cos 2 \theta + (a b \cos \phi) \sin 2 \theta = 0$

To solve this equation, the reader is referred to the results of this problem or, one can find the solution by a numerical root-finding algorithm (e.g. Newton's method).

The solution is $\theta = 0.565255 = 32.38^{\circ}$