1. Problem Q5: Two chords AB and CD intersect inside a circle at point P with $\angle ABP=40^\circ$, $\angle BPC=110^\circ$, and $\angle CDP=x$; find $x$. 2. Formula and rules: For two chords intersecting inside a circle the measure of the angle formed equals half the sum of the measures of the intercepted arcs, i.e. $\angle BPC=\tfrac{1}{2}(\text{arc }BC+\text{arc }AD)$. 3. Rule: Vertical opposite angles formed by intersecting lines are equal, so $\angle BPC=\angle APD$. 4. Rule: An inscribed angle equals half its intercepted arc, so $\angle ABP=40^\circ$ intercepts arc $AP$ and thus $\text{arc }AP=2\cdot40^\circ=80^\circ$. 5. From $\angle BPC=110^\circ$ we get $\text{arc }BC+\text{arc }AD=2\cdot110^\circ=220^\circ$. 6. The full circle measures $360^\circ$, so $\text{arc }AP+\text{arc }PC+\text{arc }CB+\text{arc }AD=360^\circ$. 7. Substitute $\text{arc }AP=80^\circ$ and $\text{arc }BC+\text{arc }AD=220^\circ$ to get $80^\circ+\text{arc }PC+220^\circ=360^\circ$. 8. Solve for $\text{arc }PC$ to obtain $\text{arc }PC=360^\circ-300^\circ=60^\circ$. 9. Angle $\angle CDP$ is an inscribed angle that intercepts arc $CP$, so $x=\tfrac{1}{2}\text{arc }CP=\tfrac{1}{2}\cdot60^\circ=30^\circ$. 10. Final answer for Q5: $x=30^\circ$, which is option (1). 1. Problem Q7: In a circle with inscribed quadrilateral ABCD the interior angle at D is $80^\circ$, and $x$ is the angle formed at B between the tangent at B and the extension of BA; find $x$. 2. Formula and rules: The angle between a tangent and a chord through the point of contact equals the inscribed angle on the opposite arc (alternate segment theorem), so the tangent-chord angle at B equals the inscribed angle that subtends the same chord $BA$. 3. In a cyclic quadrilateral opposite interior angles sum to $180^\circ$, so $\angle B+\angle D=180^\circ$ and therefore $\angle B=180^\circ-80^\circ=100^\circ$. 4. The inscribed angle that subtends chord $BA$ and lies on the opposite arc equals $\angle BDA$, and this equals the tangent-chord angle at B by the alternate segment theorem. 5. Therefore $x=\angle BDA=\angle B=100^\circ$. 6. Final answer for Q7: $x=100^\circ$, which is option (1). Final answers summary: Q5: $30^\circ$. Q7: $100^\circ$.