1. **Problem Statement:** Find the values of angles $a$, $b$, $c$, and $d$ in the circle with center $O$ where $PQ$ is tangent at $A$. Given angles: $\angle C = 110^\circ$, $\angle D = 60^\circ$. 2. **Key Theorems and Rules:** - Tangent-radius theorem: The tangent at a point on a circle is perpendicular to the radius at that point, so $\angle OAP = 90^\circ$. - Angles in the same segment are equal. - Opposite angles in cyclic quadrilaterals sum to $180^\circ$. 3. **Find angle $a$:** Since $PQ$ is tangent at $A$, $\angle OAP = 90^\circ$. Angle $a$ is $\angle OAB$, which lies on the radius $OA$ and chord $AB$. 4. **Find angle $b$:** $\angle b$ is at $B$ near center $O$. Since $O$ is center, $OB$ is radius. $\angle b$ is the angle between $OB$ and chord $BC$. 5. **Find angle $c$ and $d$:** Given $\angle C = 110^\circ$ and $\angle D = 60^\circ$. 6. **Use cyclic quadrilateral properties:** Points $A, B, C, D$ lie on the circle. Opposite angles sum to $180^\circ$. So, $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$. 7. **Calculate $\angle A$:** $$\angle A = 180^\circ - 110^\circ = 70^\circ$$ 8. **Calculate $\angle B$:** $$\angle B = 180^\circ - 60^\circ = 120^\circ$$ 9. **Relate $a$ and $b$ to $\angle A$ and $\angle B$:** Since $a$ and $b$ are angles at $A$ and $B$ near center $O$, and $O$ is center, $a$ and $b$ are central angles subtending arcs. 10. **Calculate $a$ and $b$:** Central angle equals the arc it subtends. Since $\angle A = 70^\circ$ is an inscribed angle, the central angle $a$ subtending the same arc is twice that: $$a = 2 \times 70^\circ = 140^\circ$$ Similarly for $b$: $$b = 2 \times 120^\circ = 240^\circ$$ 11. **Calculate $c$ and $d$:** Given $c = 110^\circ$, $d = 60^\circ$ (from problem statement). --- **Question 6:** 1. **Problem Statement:** Find $p$ and $q$ in the circle with given angles $45^\circ$, $65^\circ$ at $B$ and $80^\circ$ at $C$. 2. **Calculate $p$:** Angles at $B$ are $45^\circ$ and $65^\circ$, so $$p = 45^\circ + 65^\circ = 110^\circ$$ Reason: Angles on a straight line sum to $180^\circ$, and these two angles form $p$. 3. **Calculate $q$:** Sum of angles in triangle $BCD$ is $180^\circ$. Given $\angle B = 110^\circ$, $\angle C = 80^\circ$, so $$q = 180^\circ - 110^\circ - 80^\circ = -10^\circ$$ This is impossible, so re-examine. Actually, $p$ is angle at $A$, $q$ at $D$. Use alternate segment theorem or cyclic quadrilateral properties. 4. **Use cyclic quadrilateral:** Sum of opposite angles is $180^\circ$. If $p$ and $q$ are opposite angles, $$p + q = 180^\circ$$ So, $$q = 180^\circ - p = 180^\circ - 110^\circ = 70^\circ$$ --- **Final answers:** - $a = 140^\circ$ - $b = 240^\circ$ - $c = 110^\circ$ - $d = 60^\circ$ - $p = 110^\circ$ - $q = 70^\circ$