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𝒑𝒈 𝟏𝟕𝟎 − 𝟏𝟕𝟐
弧度
𝟑
𝒑𝒈 𝟏𝟕𝟎
弧度
≈ 𝟓𝟕. 𝟑𝟎 °
≈ 𝟎. 𝟎𝟏𝟕𝟒𝟓
≈ 𝟓𝟕. 𝟑𝟎 °
= 𝟖𝟕. 𝟔𝟔𝟗 °
= 𝟖𝟕 ° 𝟎. 𝟔𝟔𝟗 × 𝟔𝟎′
= 𝟏. 𝟓𝟑 × 𝟓𝟕. 𝟑𝟎 °
𝒑𝒈 𝟏𝟕𝟐
弧度 ≈ 𝟓𝟕. 𝟑𝟎 °
𝟑𝟕 ° 𝟑𝟎′ 𝟐𝟒𝟎 ° 𝟐𝟏𝟎 ° 𝟐𝟐𝟎 °
𝟕𝟔 ° 𝟏𝟐′ 𝟏𝟒𝟑 ° 𝟒𝟗′ 𝟓𝟏 ° 𝟎′ 𝟔𝟑 ° 𝟑𝟔′
𝒑𝒈 𝟏𝟕𝟏
弧度 ≈ 𝟎. 𝟎𝟏𝟕𝟒𝟓
=
𝟔𝟕. 𝟓
𝟏𝟖𝟎
𝝅 = 𝟒𝟖. 𝟖𝟓 × 𝟎. 𝟎𝟏𝟕𝟒𝟓
≈ 𝟎. 𝟎𝟏𝟕𝟒𝟓
𝒑𝒈 𝟏𝟕𝟐
弧度 ≈ 𝟎. 𝟎𝟏𝟕𝟒𝟓
𝟒𝝅
𝟑
𝟎. 𝟓𝟔 𝟑. 𝟐𝟏 𝟏. 𝟔𝟑 𝟎. 𝟕𝟐
𝟓𝝅
𝟑
𝟕𝝅
𝟑
𝝅
𝟖
𝟏𝟓𝟎 ° 𝟓𝝅
𝟔
𝟐𝟕𝟎 ° 𝟑𝝅
𝟐
𝟑𝟔𝟎 ° 𝟐𝝅 𝟔𝟑𝟎 ° 𝟕𝝅
𝟐
𝒑𝒈 𝟏𝟕𝟐 − 𝟏𝟕𝟑
𝒍
𝟑𝟔𝟎 ° = 𝟐𝝅
𝝅 =
𝟐𝟐
𝟕
= 𝟑. 𝟏𝟒𝟐
=
𝜽
𝟐𝝅
× 𝟐𝝅𝒓
𝒑𝒈 𝟏𝟕𝟑
𝒍
𝟏𝟒. 𝟑 𝒄𝒎
5.59 公尺
𝟏𝟒. 𝟗 𝒄𝒎
𝟖. 𝟓 𝒎
约 90 公圼
𝒑𝒈 𝟏𝟕𝟑
𝒍
𝒅 = 𝟏. 𝟐 𝒎
𝒍 = 𝟐𝝅𝒓 = 𝝅𝒅 = 𝟏. 𝟐 𝝅 𝒎
𝟏 𝒎𝒊𝒏 → 𝟒𝟎𝟎 𝒕𝒖𝒓𝒏𝒔
𝒍𝒎𝒊𝒏 = 𝟏. 𝟐 𝝅 × 𝟒𝟎𝟎 = 𝟒𝟖𝟎 𝝅 𝒎
𝒍𝒉𝒐𝒖𝒓 = 𝟒𝟖𝟎 𝝅 × 𝟔𝟎
= 𝟐𝟖 𝟖𝟎𝟎 𝝅 𝐦
≈ 𝟗𝟎 𝒌𝒎
𝒑𝒈 𝟏𝟕𝟒
𝒍
𝟏𝟖𝟎𝟎 °
𝟔𝝅 公尺
𝟏𝟐𝟔 公尺
𝟎. 𝟐 𝒔 𝒐𝒓 𝟏𝟐 𝒎𝒊𝒏
𝝅
𝟑
𝒎
𝟏. 𝟐 𝒓𝒂𝒅𝒊𝒂𝒏 𝟔𝟖 ° 𝟒𝟓′
𝒑𝒈 𝟏𝟕𝟒 − 𝟏𝟕𝟓
𝑺
𝑺
𝒍
𝒍 = 𝜽𝒓
𝑺 =
𝟏
𝟐
𝒍𝒓
𝒑𝒈 𝟏𝟕𝟓
𝑺
𝒍
𝑺 𝒍 = 𝜽𝒓
𝒑𝒈 𝟏𝟕𝟕
𝑺
𝑺
𝒍 = 𝜽𝒓
𝟐𝟒𝟎𝝅 𝒄𝒎𝟐
𝟏𝟑𝟓 °, (𝟑𝝅 + 𝟏𝟖) 𝒄𝒎
𝑺 =
𝟏
𝟐
𝒍𝒓
𝟐 𝒓𝒂𝒅𝒊𝒂𝒏, 𝟏𝟎𝟎 𝒄𝒎𝟐
𝟗𝟎 °
𝟖
𝟕
𝒓𝒂𝒅𝒊𝒂𝒏, 𝟏𝟏𝟐 𝒄𝒎𝟐
𝒑𝒈 𝟏𝟕𝟕
𝑺
𝑺
𝒍 = 𝜽𝒓
𝑺 =
𝟏
𝟐
𝒍𝒓
𝒓𝒔 = 𝟐 𝒓𝒄
𝑺 =
𝟏
𝟐
𝒓𝒔
𝟐𝜽 = 𝝅𝒓𝒄
𝟐
𝒍
𝒓𝒔
𝒓𝒄
radius of a sector = 2 radius of a circle
area is equal
𝟏
𝟐
× 𝟒𝒓𝒄
𝟐
𝜽 = 𝝅𝒓𝒄
𝟐
𝟐𝜽 = 𝝅
𝜽 = 𝟗𝟎 °
central angle = ?
𝒑𝒈 𝟏𝟕𝟔
𝑺
𝑺 𝒍 = 𝜽𝒓
𝒍
𝒑𝒈 𝟏𝟕𝟕
𝑺
𝑺 𝒍 = 𝜽𝒓
𝒍
𝒅
𝒑𝒈 𝟏𝟕𝟖
𝑺
𝑺
𝒍 = 𝜽𝒓
𝒍
𝟗𝟔𝝅 𝒄𝒎𝟐
𝟓𝟏𝟑. 𝟑 𝒄𝒎𝟐
, 𝟏𝟎𝟏. 𝟑 𝒄𝒎
𝟏𝟏𝟒. 𝟏𝟔 𝒄𝒎𝟐
, 𝟓𝟎. 𝟓 𝒄𝒎
𝒑𝒈 𝟏𝟕𝟖
𝑺
𝑺
𝒍 = 𝜽𝒓
𝒍
𝟏𝟏𝟒. 𝟏𝟔 𝒄𝒎𝟐
, 𝟓𝟎. 𝟓 𝒄𝒎
𝒓𝟐 = 𝑶𝑩𝟐 + 𝑶𝑪𝟐
= 𝟐𝟎𝟐
+ 𝟐𝟎𝟐
𝒓𝟐 = 𝟖𝟎𝟎
𝑺𝑶𝑪𝑫 = 𝑺𝒔𝒆𝒄𝒕𝒐𝒓 𝑩𝑪𝑫 − 𝑺∆𝑶𝑩𝑪
=
𝟏
𝟐
𝒓𝟐
𝜽 −
𝟏
𝟐
× 𝒃𝒂𝒔𝒆 × 𝒉𝒆𝒊𝒈𝒉𝒕
𝒓
𝟒𝟓 °
=
𝟏
𝟐
× 𝟖𝟎𝟎 ×
𝝅
𝟒
−
𝟏
𝟐
× 𝟐𝟎𝟐
𝟒𝟓 °
= 𝟏𝟏𝟒. 𝟏𝟔 𝒄𝒎𝟐
𝒑𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 = 𝑪𝑫 + 𝑶𝑫 + 𝑶𝑪
= 𝜽𝒓 + 𝒓 − 𝟐𝟎 + 𝟐𝟎
=
𝝅
𝟒
𝟖𝟎𝟎 + 𝟖𝟎𝟎
= 𝟖𝟎𝟎
𝝅
𝟒
+ 𝟏
= 𝟓𝟎. 𝟓 𝒄𝒎
𝒑𝒈 𝟏𝟕𝟖
𝑺
𝑺
𝒍 = 𝜽𝒓
𝒍
𝝅
𝟐
− 𝟏 𝒂𝟐
𝟒. 𝟒 𝒄𝒎, 𝟎. 𝟕 𝒄𝒎
𝒓 = 𝟏 𝒐𝒓 𝟔 , 𝒙 = 𝟏𝟐 𝒐𝒓
𝟏
𝟑
𝒑𝒈 𝟏𝟕𝟗
𝑺
𝑺
𝒍 = 𝜽𝒓
𝒍
𝟕. 𝟓 𝒄𝒎, 𝟏𝟓 𝒄𝒎, 𝟏𝟏𝟒 ° 𝟑𝟔′
𝟐. 𝟑𝟗 𝒄𝒎, 𝟕. 𝟏𝟏 𝒄𝒎

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  • 1. 𝒑𝒈 𝟏𝟕𝟎 − 𝟏𝟕𝟐 弧度 𝟑
  • 2. 𝒑𝒈 𝟏𝟕𝟎 弧度 ≈ 𝟓𝟕. 𝟑𝟎 ° ≈ 𝟎. 𝟎𝟏𝟕𝟒𝟓 ≈ 𝟓𝟕. 𝟑𝟎 ° = 𝟖𝟕. 𝟔𝟔𝟗 ° = 𝟖𝟕 ° 𝟎. 𝟔𝟔𝟗 × 𝟔𝟎′ = 𝟏. 𝟓𝟑 × 𝟓𝟕. 𝟑𝟎 °
  • 3. 𝒑𝒈 𝟏𝟕𝟐 弧度 ≈ 𝟓𝟕. 𝟑𝟎 ° 𝟑𝟕 ° 𝟑𝟎′ 𝟐𝟒𝟎 ° 𝟐𝟏𝟎 ° 𝟐𝟐𝟎 ° 𝟕𝟔 ° 𝟏𝟐′ 𝟏𝟒𝟑 ° 𝟒𝟗′ 𝟓𝟏 ° 𝟎′ 𝟔𝟑 ° 𝟑𝟔′
  • 4. 𝒑𝒈 𝟏𝟕𝟏 弧度 ≈ 𝟎. 𝟎𝟏𝟕𝟒𝟓 = 𝟔𝟕. 𝟓 𝟏𝟖𝟎 𝝅 = 𝟒𝟖. 𝟖𝟓 × 𝟎. 𝟎𝟏𝟕𝟒𝟓 ≈ 𝟎. 𝟎𝟏𝟕𝟒𝟓
  • 5. 𝒑𝒈 𝟏𝟕𝟐 弧度 ≈ 𝟎. 𝟎𝟏𝟕𝟒𝟓 𝟒𝝅 𝟑 𝟎. 𝟓𝟔 𝟑. 𝟐𝟏 𝟏. 𝟔𝟑 𝟎. 𝟕𝟐 𝟓𝝅 𝟑 𝟕𝝅 𝟑 𝝅 𝟖 𝟏𝟓𝟎 ° 𝟓𝝅 𝟔 𝟐𝟕𝟎 ° 𝟑𝝅 𝟐 𝟑𝟔𝟎 ° 𝟐𝝅 𝟔𝟑𝟎 ° 𝟕𝝅 𝟐
  • 6. 𝒑𝒈 𝟏𝟕𝟐 − 𝟏𝟕𝟑 𝒍 𝟑𝟔𝟎 ° = 𝟐𝝅 𝝅 = 𝟐𝟐 𝟕 = 𝟑. 𝟏𝟒𝟐 = 𝜽 𝟐𝝅 × 𝟐𝝅𝒓
  • 7. 𝒑𝒈 𝟏𝟕𝟑 𝒍 𝟏𝟒. 𝟑 𝒄𝒎 5.59 公尺 𝟏𝟒. 𝟗 𝒄𝒎 𝟖. 𝟓 𝒎 约 90 公圼
  • 8. 𝒑𝒈 𝟏𝟕𝟑 𝒍 𝒅 = 𝟏. 𝟐 𝒎 𝒍 = 𝟐𝝅𝒓 = 𝝅𝒅 = 𝟏. 𝟐 𝝅 𝒎 𝟏 𝒎𝒊𝒏 → 𝟒𝟎𝟎 𝒕𝒖𝒓𝒏𝒔 𝒍𝒎𝒊𝒏 = 𝟏. 𝟐 𝝅 × 𝟒𝟎𝟎 = 𝟒𝟖𝟎 𝝅 𝒎 𝒍𝒉𝒐𝒖𝒓 = 𝟒𝟖𝟎 𝝅 × 𝟔𝟎 = 𝟐𝟖 𝟖𝟎𝟎 𝝅 𝐦 ≈ 𝟗𝟎 𝒌𝒎
  • 9. 𝒑𝒈 𝟏𝟕𝟒 𝒍 𝟏𝟖𝟎𝟎 ° 𝟔𝝅 公尺 𝟏𝟐𝟔 公尺 𝟎. 𝟐 𝒔 𝒐𝒓 𝟏𝟐 𝒎𝒊𝒏 𝝅 𝟑 𝒎 𝟏. 𝟐 𝒓𝒂𝒅𝒊𝒂𝒏 𝟔𝟖 ° 𝟒𝟓′
  • 10. 𝒑𝒈 𝟏𝟕𝟒 − 𝟏𝟕𝟓 𝑺 𝑺 𝒍 𝒍 = 𝜽𝒓 𝑺 = 𝟏 𝟐 𝒍𝒓
  • 12. 𝒑𝒈 𝟏𝟕𝟕 𝑺 𝑺 𝒍 = 𝜽𝒓 𝟐𝟒𝟎𝝅 𝒄𝒎𝟐 𝟏𝟑𝟓 °, (𝟑𝝅 + 𝟏𝟖) 𝒄𝒎 𝑺 = 𝟏 𝟐 𝒍𝒓 𝟐 𝒓𝒂𝒅𝒊𝒂𝒏, 𝟏𝟎𝟎 𝒄𝒎𝟐 𝟗𝟎 ° 𝟖 𝟕 𝒓𝒂𝒅𝒊𝒂𝒏, 𝟏𝟏𝟐 𝒄𝒎𝟐
  • 13. 𝒑𝒈 𝟏𝟕𝟕 𝑺 𝑺 𝒍 = 𝜽𝒓 𝑺 = 𝟏 𝟐 𝒍𝒓 𝒓𝒔 = 𝟐 𝒓𝒄 𝑺 = 𝟏 𝟐 𝒓𝒔 𝟐𝜽 = 𝝅𝒓𝒄 𝟐 𝒍 𝒓𝒔 𝒓𝒄 radius of a sector = 2 radius of a circle area is equal 𝟏 𝟐 × 𝟒𝒓𝒄 𝟐 𝜽 = 𝝅𝒓𝒄 𝟐 𝟐𝜽 = 𝝅 𝜽 = 𝟗𝟎 ° central angle = ?
  • 15. 𝒑𝒈 𝟏𝟕𝟕 𝑺 𝑺 𝒍 = 𝜽𝒓 𝒍 𝒅
  • 16. 𝒑𝒈 𝟏𝟕𝟖 𝑺 𝑺 𝒍 = 𝜽𝒓 𝒍 𝟗𝟔𝝅 𝒄𝒎𝟐 𝟓𝟏𝟑. 𝟑 𝒄𝒎𝟐 , 𝟏𝟎𝟏. 𝟑 𝒄𝒎 𝟏𝟏𝟒. 𝟏𝟔 𝒄𝒎𝟐 , 𝟓𝟎. 𝟓 𝒄𝒎
  • 17. 𝒑𝒈 𝟏𝟕𝟖 𝑺 𝑺 𝒍 = 𝜽𝒓 𝒍 𝟏𝟏𝟒. 𝟏𝟔 𝒄𝒎𝟐 , 𝟓𝟎. 𝟓 𝒄𝒎 𝒓𝟐 = 𝑶𝑩𝟐 + 𝑶𝑪𝟐 = 𝟐𝟎𝟐 + 𝟐𝟎𝟐 𝒓𝟐 = 𝟖𝟎𝟎 𝑺𝑶𝑪𝑫 = 𝑺𝒔𝒆𝒄𝒕𝒐𝒓 𝑩𝑪𝑫 − 𝑺∆𝑶𝑩𝑪 = 𝟏 𝟐 𝒓𝟐 𝜽 − 𝟏 𝟐 × 𝒃𝒂𝒔𝒆 × 𝒉𝒆𝒊𝒈𝒉𝒕 𝒓 𝟒𝟓 ° = 𝟏 𝟐 × 𝟖𝟎𝟎 × 𝝅 𝟒 − 𝟏 𝟐 × 𝟐𝟎𝟐 𝟒𝟓 ° = 𝟏𝟏𝟒. 𝟏𝟔 𝒄𝒎𝟐 𝒑𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 = 𝑪𝑫 + 𝑶𝑫 + 𝑶𝑪 = 𝜽𝒓 + 𝒓 − 𝟐𝟎 + 𝟐𝟎 = 𝝅 𝟒 𝟖𝟎𝟎 + 𝟖𝟎𝟎 = 𝟖𝟎𝟎 𝝅 𝟒 + 𝟏 = 𝟓𝟎. 𝟓 𝒄𝒎
  • 18. 𝒑𝒈 𝟏𝟕𝟖 𝑺 𝑺 𝒍 = 𝜽𝒓 𝒍 𝝅 𝟐 − 𝟏 𝒂𝟐 𝟒. 𝟒 𝒄𝒎, 𝟎. 𝟕 𝒄𝒎 𝒓 = 𝟏 𝒐𝒓 𝟔 , 𝒙 = 𝟏𝟐 𝒐𝒓 𝟏 𝟑
  • 19. 𝒑𝒈 𝟏𝟕𝟗 𝑺 𝑺 𝒍 = 𝜽𝒓 𝒍 𝟕. 𝟓 𝒄𝒎, 𝟏𝟓 𝒄𝒎, 𝟏𝟏𝟒 ° 𝟑𝟔′ 𝟐. 𝟑𝟗 𝒄𝒎, 𝟕. 𝟏𝟏 𝒄𝒎

Editor's Notes

  1. https://www.onlinemathlearning.com/degrees-radians-3.html The ratio of arc length to radius is 3:1, and the central angle to this arc radian value?
  2. https://www.onlinemathlearning.com/degrees-radians-3.html
  3. https://www.onlinemathlearning.com/arc-length.html 5. The central angle is 2/3 of the circumferential angle. 6. one clock needle is 7 cm long, find 20 minutes after, the distance moved by its tip?
  4. The diameter of the driving wheel = 1.2 m Driving wheel rotates 400 turn per minute, the train travels how many kilometers per hour?
  5. 6. The diameter of the flywheel is 1.2 km, if it rotates 300 times per minute (a) the arc length that a point on the circle of the flywheel turns per second (c) the time required for one rotation of the flywheel 7. Someone is running at a constant speed along a circular track, passing the central angle of the arc is 2 6/7 radian every minute, if the person ran a total of 5250 km in 14 minutes and 40 seconds, find radius. 8. Arc AB = 1/4*O1A/2, Arc AC = 1/8*O1A 10. radius = 120 mm, arc = 144 mm, find theta in radian and degree.
  6. https://www.onlinemathlearning.com/area-sector.html
  7. 5. 公分= cm, perimeter (sector) = 0.5 perimeter (circle)
  8. The radius OC and the diameter AB are perpendicular to each other, with B as the center, BC as the radius, and the arc intersects AB and D. If the radius of the circle ABC is 20 cm, find shaded are and perimeter.
  9. draw a semicircle in the square with each side as the diameter, and find the enclosed area of ​​(the shadow part) 8*(90 degree S-triangle)
  10. 13. prove that the ratio of the area of ​​the two sectors with the same central angle is equal to the ratio of the square of the radius