1. **Stating the problem:** We have a circle with points W, X, Y, Z on its circumference forming four triangles: WXY, WYZ, XYZ, and WZY. Angles at these points are given or labeled as variables: at W angles $a^\circ$ and $b^\circ$, at X angles $70^\circ$ and $28^\circ$, at Y angles $c^\circ$ and $d^\circ$, and at Z angles $31^\circ$ and $51^\circ$. We want to find the values of $a$, $b$, $c$, and $d$.\n\n2. **Key rule:** The sum of angles in any triangle is $180^\circ$. Also, angles subtended by the same chord in a circle are equal, and the sum of opposite angles in a cyclic quadrilateral is $180^\circ$.\n\n3. **Triangle WXY:** Angles are $a^\circ$ (at W), $70^\circ$ (at X), and $b^\circ$ (at Y). Using the triangle sum rule: $$a + 70 + b = 180$$\n\n4. **Triangle WYZ:** Angles are $b^\circ$ (at W), $c^\circ$ (at Y), and $31^\circ$ (at Z). Sum: $$b + c + 31 = 180$$\n\n5. **Triangle XYZ:** Angles are $28^\circ$ (at X), $c^\circ$ (at Y), and $51^\circ$ (at Z). Sum: $$28 + c + 51 = 180$$\n\n6. **Triangle WZY:** Angles are $a^\circ$ (at W), $d^\circ$ (at Y), and $51^\circ$ (at Z). Sum: $$a + d + 51 = 180$$\n\n7. **Solve for $c$ from triangle XYZ:** $$28 + c + 51 = 180 \Rightarrow c = 180 - 79 = 101$$\n\n8. **Solve for $b$ from triangle WYZ:** $$b + 101 + 31 = 180 \Rightarrow b = 180 - 132 = 48$$\n\n9. **Solve for $a$ from triangle WXY:** $$a + 70 + 48 = 180 \Rightarrow a = 180 - 118 = 62$$\n\n10. **Solve for $d$ from triangle WZY:** $$62 + d + 51 = 180 \Rightarrow d = 180 - 113 = 67$$\n\n**Final answers:** $a = 62^\circ$, $b = 48^\circ$, $c = 101^\circ$, $d = 67^\circ$.