1. **Problem Statement:** We have a circle with points A, B, D, and E on its circumference and center O. EB is a diameter, BC is tangent at B, angle EBD = 30°, and angle ACB = 20°. F is the intersection of BE and AD. We need to find: (i) angle ADB (ii) angle BAC 2. **Key Properties and Formulas:** - Angle subtended by a diameter is a right angle: $$\angle AEB = 90^\circ$$ - Tangent-radius theorem: tangent at a point is perpendicular to the radius at that point. - Angles subtended by the same chord are equal. - Exterior angle theorem in triangles. 3. **Find angle ADB:** - Since EB is diameter, $$\angle EAB = 90^\circ$$ (angle in semicircle). - Given $$\angle EBD = 30^\circ$$. - Triangle EBD lies on the circle, so $$\angle EBD$$ subtends arc ED. - Angle ADB subtends the same arc ED as angle EBD but from the opposite side of the chord. - By the circle theorem, angles subtended by the same chord are equal, so: $$\angle ADB = \angle EBD = 30^\circ$$ 4. **Find angle BAC:** - Given $$\angle ACB = 20^\circ$$ and BC is tangent at B. - Tangent-chord angle theorem states that the angle between tangent and chord equals the angle in the alternate segment. - So, $$\angle ABC = \angle ACB = 20^\circ$$. - In triangle ABC, sum of angles is 180°: $$\angle BAC + \angle ABC + \angle ACB = 180^\circ$$ $$\angle BAC + 20^\circ + 20^\circ = 180^\circ$$ $$\angle BAC = 180^\circ - 40^\circ = 140^\circ$$ **Final answers:** (i) $$\angle ADB = 30^\circ$$ (ii) $$\angle BAC = 140^\circ$$