1. **Problem statement:** Given a circle with center $O$, diameter $AB$, tangent line $ED$ at $C$, angle $\angle A = 36^\circ$, $AO = 12$, and chord $CD = 9$, solve for various angles and lengths as well as the area of the circle. 2. Since $AB$ is a diameter, $AO = OB = 12$, so radius $r = 12$. 3. a) Find $m \angle BC$: - Note that point $C$ lies on the circle, forming triangle $ABC$. - $\angle A = 36^\circ$ given. - Since $AB$ is diameter, $\angle ACB = 90^\circ$ by Thales' theorem. - Using the fact that angles in triangle $ABC$ sum to $180^\circ$, $$m \angle BC + m \angle AC + m \angle A = 180^\circ,$$ and since $m \angle A = 36^\circ$ and $m \angle C = 90^\circ$, $$m \angle BC = 180^\circ - 36^\circ - 90^\circ = 54^\circ.$$ 4. b) $m \angle AC$ is the angle at $A$ subtended by chord $AC$: - From triangle $ABC$, we get $m \angle AC = 90^\circ$ because $\angle C$ is a right angle at the circumference opposite to diameter $AB$. 5. c) Find $m \angle OCE$: - Since $ED$ is tangent at point $C$, $\angle OCE$ is the angle between radius $OC$ and tangent $ED$. - The tangent is perpendicular to radius at point of tangency, so $$m \angle OCE = 90^\circ.$$ 6. d) Find $m \angle AOC$: - $\triangle AOC$ is isosceles with $AO = OC = 12$ (radii). - Use $m \angle A = 36^\circ$ which subtends arc $AC$, so central angle $m \angle AOC = 2 \times m \angle AC = 2 \times 36^\circ = 72^\circ$ by the inscribed angle theorem. 7. e) Find $m \angle CDB$: - $D$ lies on line through $B$ and $CD = 9$. - Since $AB$ is diameter and $C$ is on circle, $\angle CDB$ subtends chord $CB$ and is equal to $\angle CAB= 36^\circ$ because angles subtending same chord are equal. 8. f) Find $m \angle CBD$: - In triangle $CBD$, sum of angles is $180^\circ$. - $m \angle CDB = 36^\circ$ from (e). - $m \angle CBD$ is complementary angle to $m \angle BCD$, but without extra info, assume $$m \angle CBD = 54^\circ$$ because $36^\circ + 54^\circ + 90^\circ =180^\circ$ where $m \angle BCD$ is right angle from tangent property. 9. g) Find $m \angle DCB$: - This is angle between chord $CD$ and $CB$ at $C$. - Since $ED$ tangent at $C$ and $CD$ chord, $m \angle DCB = m \angle CBD = 54^\circ$. 10. h) Find length $OC$: - Radius $OC = 12$ because $C$ lies on circle. 11. i) Find length $BD$: - Since $AB = 24$, and $CD = 9$, use Pythagorean or segment properties. - $BD = BC + CD$. - $BC$ can be found by law of cosines in $\triangle BOC$ or by chord length formula. - Using law of cosines in $\triangle BOC$ with $OB=12$, $OC=12$, $m \angle BOC = 2 \times 54^\circ = 108^\circ$ (central angle opposite arc $BC$), $$BC=\sqrt{12^2 + 12^2 - 2\times12\times12\times\cos 108^\circ} = \sqrt{288-288\times \cos 108^\circ}.$$ - $\cos 108^\circ = -\cos 72^\circ = -0.3090$, so $$BC = \sqrt{288 + 288 \times 0.3090} = \sqrt{288 + 89} = \sqrt{377} = \frac{\sqrt{377}}{1}.$$ - Thus, $$BD = BC + CD = \sqrt{377} + 9.$$ 12. j) Area of $\odot O$: - Area formula is $$\pi r^2 = \pi \times 12^2 = 144 \pi.$$ **Final answers:** - a) $m \angle BC = 54^\circ$ - b) $m \angle AC = 90^\circ$ - c) $m \angle OCE = 90^\circ$ - d) $m \angle AOC = 72^\circ$ - e) $m \angle CDB = 36^\circ$ - f) $m \angle CBD = 54^\circ$ - g) $m \angle DCB = 54^\circ$ - h) $OC = 12$ - i) $BD = 9 + \sqrt{377}$ - j) Area $= 144\pi$