1. Problem: Given $m\angle COE = 40^\circ$, find arc $CE$. Formula: The measure of an arc is equal to the measure of its central angle. Since $m\angle COE = 40^\circ$, arc $CE = 40^\circ$. 2. Problem: Given arc $AC = 85^\circ$, find $m\angle CAO$. Rule: An inscribed angle is half the measure of its intercepted arc. So, $m\angle CAO = \frac{1}{2} \times 85^\circ = 42.5^\circ$. 3. Problem: Given arc $AG = 120^\circ$, find $m\angle GHA$. Rule: Inscribed angle $m\angle GHA = \frac{1}{2} \times$ arc $AG = \frac{1}{2} \times 120^\circ = 60^\circ$. 4. Problem: Given $m\angle AHG = 75^\circ$, find arc $GA$. Rule: Arc $GA = 2 \times m\angle AHG = 2 \times 75^\circ = 150^\circ$. 5. Problem: Given $m\angle GHA = 30^\circ$ and arc $CE = 45^\circ$, find sum of arc $GA$ and $m\angle EOC$. Rule: $m\angle GHA = \frac{1}{2} \times$ arc $GA$ so arc $GA = 2 \times 30^\circ = 60^\circ$. Central angle $m\angle EOC = $ arc $CE = 45^\circ$. Sum = $60^\circ + 45^\circ = 105^\circ$. 6. Problem: Find $m\angle E$ formed by intersecting secants/tangents outside the circle. Rule: $m\angle E = \frac{1}{2} (\text{difference of intercepted arcs})$. Given arcs 130° and 60°, so $m\angle E = \frac{1}{2} (130^\circ - 60^\circ) = 35^\circ$. 7-8. Problem: Given arc $NP = 30^\circ$ and $m\angle O = 45^\circ$, find arc $QM$. Rule: For intersecting secants, $m\angle O = \frac{1}{2} (\text{sum of arcs } NP + QM)$. So, $45^\circ = \frac{1}{2} (30^\circ + QM)$ implies $90^\circ = 30^\circ + QM$ so $QM = 60^\circ$. 9-10. Problem: Given $VW=14$ cm, $WX=4$ cm, $XY=7$ cm, find $XZ$. Rule: For two secants intersecting outside the circle, $VW \times WX = XY \times XZ$. Calculate: $14 \times 4 = 7 \times XZ$ so $56 = 7 \times XZ$ implies $XZ = 8$ cm. 11-12. Problem: Given $DE=9$ cm, $CE=12$ cm, $AB=6$ cm, find $AE$. Rule: For two secants intersecting outside the circle, $DE \times CE = AE \times AB$. Calculate: $9 \times 12 = AE \times 6$ so $108 = 6 \times AE$ implies $AE = 18$ cm. 13. Problem: Given $AC=18$ m, $BC=8$ m, find $DC$. Rule: For two secants intersecting outside the circle, $AC \times BC = DC \times BC$ (assuming typo, likely $DC \times BC$ or $DC \times something$). Assuming $AC \times BC = DC \times BC$ is a typo, likely $AC \times BC = DC \times BC$ is incorrect. Without more info, cannot solve. 14. Problem: Given $DC=9$ cm, $BC=3$ cm, find $AC$. Same issue as above, insufficient info to solve. 15. Problem: Given $AC=25$ m, $DC=15$ m, find $BC$. Same issue as above, insufficient info to solve. Final answers summarized: 1. $40^\circ$ 2. $42.5^\circ$ 3. $60^\circ$ 4. $150^\circ$ 5. $105^\circ$ 6. $35^\circ$ 7-8. $60^\circ$ 9-10. $8$ cm 11-12. $18$ cm 13-15. Insufficient data to solve.