1. **Statement of the problem:** We are given four geometric setups involving angles labeled with expressions in terms of $x$, $y$, $z$, and some numerical angles. We need to find the value of $k$ in each case. --- **Top-left circle with triangle inside:** 2. Given angles: $3x^\circ$, $y^\circ$, $z^\circ$, and two angles labeled $x^\circ$ on the edges. 3. Equation given: $k = z - (x + y)$. 4. We need more info to find $k$, but we will treat it as is since no values are provided. --- **Top-right circle with triangle:** 5. Given angles: $130^\circ$, $y^\circ$, $70^\circ$, $x^\circ$. 6. Equation: $k = 2(y - x) + 45^\circ$. 7. Since sum of angles in a triangle equals $180^\circ$, we have: $$130 + y + 70 = 180 \implies y = 180 - 200 = -20^\circ,$$ which is impossible; likely an error or more info is needed. --- **Bottom-left circle with radius OA and tangent AT:** 8. Given: $\angle OAB = 68^\circ$, $\angle ABO = 20^\circ$, $k = \angle ATB$. 9. Triangle $OAB$ has angles summing to $180^\circ$, so: $$\angle AOB = 180 - 68 - 20 = 92^\circ.$$ 10. Tangent-radius property states $\angle OAT = 90^\circ$, since $AT$ is tangent to circle at $A$. 11. To find $k = \angle ATB$, note that $k$ is angle between tangent $AT$ and chord $BT$; properties of circle and tangent give $k = \angle OBA = 20^\circ$. --- **Bottom-right cyclic quadrilateral ABCD:** 12. Given $AD \parallel BC$, $\angle DCB = 65^\circ$, and $k = \angle CBE$. 13. Since $AD \parallel BC$, alternate interior angles at $C$ and $B$ are equal, so: $$k = \angle CBE = \angle DCB = 65^\circ.$$ --- **Final answers:** - For bottom-left figure: $k = 20^\circ$ - For bottom-right figure: $k = 65^\circ$ - Top-left and top-right lack sufficient info to determine numerical $k$.