Решение:
Для решения задачи будем использовать данные с транспортира. Угол измеряется от луча OA.
- \( \angle AOD = 30^{\circ} \)
- \( \angle AOC = 40^{\circ} \)
- \( \angle BOD = 10^{\circ} \) (разница между \( \angle AOC \) и \( \angle BOC \))
- \( \angle AOT = 120^{\circ} \)
- \( \angle AOP = 150^{\circ} \)
- \( \angle VOZ = 180^{\circ} \) (развернутый угол)
- \( \angle VOX = 170^{\circ} \)
- \( \angle VOL = 160^{\circ} \)
- \( \angle VOF = 170^{\circ} \)
- \( \angle AOZ = 180^{\circ} \)
- \( \angle AOH = 90^{\circ} \)
- \( \angle VOH = 180^{\circ} - 90^{\circ} = 90^{\circ} \)
- \( \angle VOL = 160^{\circ} \)
- \( \angle AOE = 60^{\circ} \)
- \( \angle AOM = 140^{\circ} \)
- \( \angle VON = 150^{\circ} \)
- \( \angle COD = 10^{\circ} \) (разница между \( \angle AOD \) и \( \angle AOC \))
- \( \angle DOE = 20^{\circ} \) (разница между \( \angle AOE \) и \( \angle AOD \))
- \( \angle EOS = 70^{\circ} \) (разница между \( \angle AOS \) и \( \angle AOE \))
- \( \angle FOT = 120^{\circ} - 110^{\circ} = 10^{\circ} \)
- \( \angle EOT = 120^{\circ} - 60^{\circ} = 60^{\circ} \)
- \( \angle KOS = 150^{\circ} - 140^{\circ} = 10^{\circ} \)
- \( \angle SOP = 180^{\circ} - 150^{\circ} = 30^{\circ} \)
- \( \angle MOC = 140^{\circ} - 40^{\circ} = 100^{\circ} \)
- \( \angle NOL = 150^{\circ} - 110^{\circ} = 40^{\circ} \)
- \( \angle ZOT = 120^{\circ} - 180^{\circ} = -60^{\circ} \) (или \( 60^{\circ} \) по модулю)
- \( \angle KOV = 160^{\circ} - 150^{\circ} = 10^{\circ} \)
- \( \angle LOC = 160^{\circ} - 40^{\circ} = 120^{\circ} \)
- \( \angle XOL = 150^{\circ} - 130^{\circ} = 20^{\circ} \)
- \( \angle NOC = 150^{\circ} - 110^{\circ} = 40^{\circ} \)
Ответ: 1) 30° 2) 40° 3) 10° 4) 120° 5) 150° 6) 180° 7) 170° 8) 160° 9) 170° 10) 180° 11) 90° 12) 90° 13) 160° 14) 60° 15) 140° 16) 150° 17) 10° 18) 20° 19) 70° 20) 10° 21) 60° 22) 10° 23) 30° 24) 100° 25) 40° 26) 60° 27) 10° 28) 120° 29) 20° 30) 40°