1. **Statement of the first problem:** At 1 hour 20 minutes, find the angle between the hour hand and the minute hand of a clock. 2. **Calculate the positions of the hands:** - The hour hand moves 30 degrees per hour plus 0.5 degrees per minute. - At 1 hour 20 minutes: hour hand angle = $$30 \times 1 + 0.5 \times 20 = 30 + 10 = 40$$ degrees from 12 o'clock. - The minute hand moves 6 degrees per minute. - At 20 minutes: minute hand angle = $$6 \times 20 = 120$$ degrees from 12 o'clock. 3. **Find the angle between the two hands:** - Angle = absolute difference = $$|120 - 40| = 80$$ degrees. - The angle between clock hands is always the smaller of the two possible angles. - Since 80 degrees is less than 180 degrees, the answer is $$80$$ degrees. --- 4. **Statement of the second problem:** In isosceles triangle ABC, AC is the diameter of the circle centered at O. Point D lies on AC such that AD:DB=? Find angle $$\angle AOD$$. 5. **Analyze given information:** - AC is diameter, so O is midpoint of AC. - Thus, AO = OC = half of AC. - Since AC is diameter, the triangle ABC inscribed in circle with hypotenuse AC is right angled at B. 6. **Find ratio AD:DB:** - Since O is midpoint of AC: AO = OC. - Point D is on AC; suppose D coincides with O (as no other info). - So AD:DB = AO:OC = 1:1. 7. **Calculate angle $$\angle AOD$$:** - Points A, O, D are collinear if D is on line AC. - Assuming D = O, $$\angle AOD = 0$$ degrees. - If D is distinct, additional info is needed. **Final answers:** - Angle between clock hands: $$80$$ degrees. - Ratio $$AD:DB = 1:1$$. - Angle $$\angle AOD = 0$$ degrees (if D coincides with O).