Find the yellow angle?

Let the center of the quarter circle be $C$ and its radius be 1. Let the rays tangent to the circle at $A$ and $B$ . Then the perpendiculars at $A$ and $B$ meet at the center $O$ of the circle. Then $OA = OB = r$ , the radius of the circle. Let $\angle ACB = \theta$ . Let $OD$ be perpendicular to the right side of the square then:

$\begin{aligned} OD + OC \times \cos 45^\circ & = 1 \\ r + \frac {\cos 45^\circ}{\cos \frac \theta 2} & = 1 \\ \tan \frac \theta 2 + \frac 1{\sqrt 2 \cos \frac \theta 2} & = 1 & \small \blue{\text{Multiply both sides by }\sqrt 2\cos \frac \theta 2} \\ \sqrt 2 \sin \frac \theta 2 + 1 & = \sqrt 2 \cos \frac \theta 2 \\ 2 \left(\frac 1{\sqrt 2}\cos \frac \theta 2 - \frac 1{\sqrt 2} \sin \frac \theta 2 \right) & = 1 \\ \cos \left(\frac \theta 2 + 45^\circ \right) & = \frac 12 \\ \frac \theta 2 + 45^\circ & = 60^\circ \\ \implies \theta & = \boxed{30^\circ} \end{aligned}$

Let the radius of the quarter circle be $a$ and that of the full circle be $b$ . Then $a^2+b^2=2(a-b)^2$ or $a^2-4ab+b^2=0$ . This yields $\dfrac{b}{a}=2-√3$ (since $b<a$ ). Thus half of the required angle is $\tan^{-1} (\dfrac{b}{a})=\tan^{-1} {(2-√3)}=15\degree$ and hence the required angle is $\boxed {30\degree}$