1. Problem statement: We have a cyclic quadrilateral WXYZ (points on a circle) with diagonals WY and XZ intersecting inside, and the given small angles at the vertices are: at X: 70° and 28°; at Z: 31° and 51°; at W: a° and b°; at Y: c° and d°. Find $a,b,c,d$. 2. Plan and theorem: Use the inscribed-angle theorem: an inscribed angle equals half the measure of its intercepted arc. 3. Parameterize arcs: Let $A,B,C,D$ be the measures of arcs $WX,XY,YZ,ZW$ respectively, measured in degrees, so $A+B+C+D=360$. 4. Express each given small inscribed angle as half the intercepted arc and solve for arcs. $$\text{WXZ}=\frac{D}{2}=70$$ Therefore $$D=140$$ $$\text{ZXY}=\frac{C}{2}=28$$ Therefore $$C=56$$ $$\text{WZX}=\frac{A}{2}=51$$ Therefore $$A=102$$ From $A+B+C+D=360$ we get $$B=360-(A+C+D)=360-(102+56+140)=62$$ So $$A=102,\;B=62,\;C=56,\;D=140$$ 5. Compute the small angles at W and Y from the appropriate intercepted arcs. At W the two angles are $$\text{XWY}=\frac{B}{2}=\frac{62}{2}=31$$ and $$\text{YWZ}=\frac{C}{2}=\frac{56}{2}=28$$ Thus the two angles at W are $31^\circ$ and $28^\circ$. At Y the two angles are $$\text{XYW}=\frac{A}{2}=\frac{102}{2}=51$$ and $$\text{WYZ}=\frac{D}{2}=\frac{140}{2}=70$$ Thus the two angles at Y are $51^\circ$ and $70^\circ$. 6. Assigning labels to match the diagram (first listed angle at W is $a$, second is $b$, first at Y is $c$, second is $d$) gives the final answers. Final answer: $a=31^\circ$, $b=28^\circ$, $c=51^\circ$, $d=70^\circ$.