Pre RMO

Let the smaller circle has center $O'$ and let that circle be tangent to $XY$ at a point $M$ .

Const:- Join $OO'$ , $O'M$ and $O'Z..$

Clearly, $\angle O'ZP = \angle O'PZ = \angle OPY =\angle OYP$ .

=> $\angle O'ZP = \angle OYP$

=> $O'Z || OY$

=> $\dfrac{O'P}{O'O} = \dfrac{PZ}{YZ} = \dfrac{1}{2}$

=> $\dfrac{O'M}{O'O} = \dfrac{1}{2}$

=> $\angle O'OM = 30^{\circ}$

=> $\angle PYX = (\angle O'OM /2) = 15^{\circ}$

K.I.P.K.I.G

Thanks NIRANJAN for the solution but I cannot understand your images. They seem like cartoon pictures.

$Fig~. IV:- ~~Let ~YP ~ = ~6n.\\ Fig. ~II:- ~~\text{Let Q be the center and r the radius of the small circle, R radius of the big circle.}\\ \text{The point of tangency to YX is C. Small circle cut OP at D. }\\ Fig. ~III:- ~~OB, ~DZ, ~and ~ QA ~are \bot s ~on ~YP ~ and ~\therefore ~all ~ parallel. \color{#3D99F6}{*}\\ Given~YZ=\dfrac 2 3 *YP=4n, ~also ~YB~\bot ~to ~chord ~YP~\implies~YB=\dfrac {YP} 2=3n, YB~\bot ~to ~chord ~YP~\implies ~BZ=n.\\ Given~ZP=\dfrac 1 3 *YP=2n. ~~But ~QZ ~is ~chord ~bisector.~ So ~ZA=AP=n.\\ \therefore~in ~\Delta ~POB, A, ~Z, ~B ~trisects ~BP.\\ QA | | DZ | | OB, \color{#3D99F6}{*},~ ~\implies ~Q,~D,~O, ~also~trisects ~OP.\\ \therefore ~OD=r.~ But ~OD=R - 2r ~ALSO. ~~~~~\therefore ~R-2r=r ~ \implies~ R=3r. \\ Fig. ~I:-\therefore ~in ~rt. \angle ed \Delta ~COQ, ~hypotinous ~OQ=2*legCQ.\\ \implies~ \angle~ COQ=30^o~ at~ the~ center ~ O.\\ \therefore ~on ~ the~circumferance ~\angle XYP=\dfrac 1 2*30=\Large ~~~\color{#D61F06}{15^o}$