A geometry problem by Arpit Sah

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Let $\angle BOC = \theta$

$r\theta = \frac{20\pi}{3} \longrightarrow 10\theta = \frac{20\pi}{3} \longrightarrow \theta = \frac{2\pi}{3} = 120^\circ$

$\angle AOC = \angle ACO = \angle CAO = 60^\circ$

$\angle ACD = \angle BCD = 45^\circ$

So, $\angle OCD = 15^\circ$ and $\angle COD = 150^\circ$

$$

$\cos^2 15^\circ = \frac{1+\cos 30^\circ}{2}$

$\cos^2 15^\circ = \frac{1+\frac{\sqrt{3}}{2}}{2}$

$\cos^2 15^\circ = \frac{2+\sqrt{3}}{4}$

$cos^2 15^\circ = \frac{8+4\sqrt{3}}{16}$

$cos^2 15^\circ = \frac{6+2\sqrt{12} + 2}{16}$

$cos^2 15^\circ = \frac{(\sqrt{6} + \sqrt{2})^2 }{16}$

$cos 15^\circ = \frac{\sqrt{6} + \sqrt{2} }{4}$

Now Using Sine Law in $\triangle COD$ ,

$\frac{CD}{\sin 150^\circ} = \frac{CO}{\sin 15^\circ}$

$CD = \frac{\sin 30^\circ \times 10}{\sin 15^\circ}$

$CD = \frac{2\sin 15^\circ \cos 15^\circ \times 10}{\sin 15^\circ}$

$CD = 2 \cos 15^\circ \times 10$

$CD = 5\sqrt{2} (\sqrt{3}+1)$