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SUEC 高中 Adv Maths (Triangulation Problem)
- 1. 𝒑𝒈 𝟐𝟐𝟏 − 𝟐𝟐𝟐 ?
- 2. 𝒑𝒈 𝟐𝟐𝟐 − 𝟐𝟐𝟑 𝑨𝑫 =?
- 3. 𝟐𝟎𝟎. 𝟔 𝒎 𝟒𝟒. 𝟔 𝒄𝒎 𝒑𝒈 𝟐𝟐𝟓
- 4. 𝟒𝟒. 𝟔 𝒄𝒎 Q P B O A 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑖𝑠 𝑓𝑙𝑎𝑡, 𝑂𝑄 = 𝑂𝐴 + 𝐴𝑃 25 125 = 25 + 125 = 150 𝑐𝑚 𝑄𝑃 = 𝑂𝑄 − 𝑃 ? ∆ 𝐴𝑂𝑃 𝑐2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 𝑐𝑜𝑠 𝐶 𝐴𝑃 2 = (𝑂𝐴)2 +(𝑂𝑃)2 − 2 𝑂𝐴 𝑂𝑃 𝑐𝑜𝑠 135 ° 125 2 = 25 2 + 𝑂𝑃 2 − 2 25 𝑂𝑃 𝑐𝑜𝑠 135 ° = 106.07 cm 𝑂𝑃 2 + 25 2 OP − 15,000 = 0 𝑥 = −𝑏 ± 𝑏2 − 4𝑎𝑐 2𝑎 𝑂𝑃 = −25 2 ± 175 2 2 𝑄𝑃 = 𝑂𝑄 − 𝑂𝑃 = 150 − 106.07 = 43.93 𝑐𝑚
- 5. 𝟏𝟒. 𝟒𝟐 𝒑𝒈 𝟐𝟐𝟓 𝟐𝟑𝟗. 𝟓 𝒎
- 6. 𝑨𝑩 = 𝟏𝟐 𝒏𝒎 𝒔𝒉𝒊𝒑 𝑨 = 𝟒 𝒏𝒎/𝒉 𝒔𝒉𝒊𝒑 𝑩 = 𝟔 𝒏𝒎/𝒉 𝒂𝒇𝒕𝒆𝒓 𝟐 𝒉𝒓𝒔 → 𝟖 𝒏𝒎 𝒂𝒇𝒕𝒆𝒓 𝟐 𝒉𝒓𝒔 → 𝟏𝟐 𝒏𝒎 = 𝑩𝑫 = 𝑨𝑩 𝑩𝑬 = 𝑨𝑩 − 𝟖 𝒏𝒎 = 𝟒 𝒏𝒎 ? ∆ 𝐵𝐷𝐸 𝑐2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 𝑐𝑜𝑠 𝐶 𝐶𝐷 2 = (𝐵𝐸)2 +(𝐵𝐷)2 − 2 𝐵𝐶 𝐵𝐷 𝑐𝑜𝑠 120 ° = 4 2 + 12 2 − 2 4 12 𝑐𝑜𝑠 120 ° E = 208 ∴ 𝐶𝐷 = 14.42 𝑛𝑚
- 7. 𝒑𝒈 𝟐𝟐𝟑
- 8. 𝒑𝒈 𝟐𝟐𝟒 ?
- 9. 𝒑𝒈 𝟐𝟐𝟓 𝟏𝟎. 𝟑𝟔𝟔 𝒌𝒎 𝒂𝒏𝒈𝒍𝒆 𝑨𝑪𝑩 = 𝟐𝟏 ° 𝑨𝑪 = 𝟏𝟎. 𝟗𝟐 𝒌𝒎 𝟑𝟑𝟔 ° 𝑩𝑫 = 𝟏𝟒. 𝟑𝟏 𝒌𝒎
- 10. 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒐𝒇 𝑪 𝒇𝒓𝒐𝒎 𝑫 = 𝟑𝟔𝟎 ° − 𝟐𝟒 ° ∆ 𝐴𝐶𝐷 𝑐2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 𝑐𝑜𝑠 𝐶 𝐶𝐷 2 = (𝐴𝐶)2 +(𝐴𝐷)2 − 2 𝐴𝐶 𝐴𝐷 𝑐𝑜𝑠 145 ° = 17 2 + 8 2 − 2 17 8 𝑐𝑜𝑠 145 ° 𝐶𝐷 = 24 𝑘𝑚 = 575.81 𝑐𝑜𝑠 𝐴 = 𝑏2 + 𝑐2 − 𝑎2 2𝑏𝑐 = 8 2 + 24 2 − 17 2 2 8 24 angle ADC 𝑎𝑛𝑔𝑙𝑒 𝐴𝐷𝐶 = 24 ° = 0.914 = 𝟑𝟑𝟔 ° ∆ 𝐴𝐵𝐷 𝑐2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 𝑐𝑜𝑠 𝐶 𝐵𝐷 2 = (𝐴𝐵)2 +(𝐴𝐷)2 − 2 𝐴𝐵 𝐴𝐷 𝑐𝑜𝑠 145 ° = 7 2 + 8 2 − 2 7 8 𝑐𝑜𝑠 145 ° ∴ 𝐵𝐷 = 14.31 𝑘𝑚 = 204,75
- 11. 𝒑𝒈 𝟏𝟕𝟎
- 12. 𝒑𝒈 𝟐𝟐𝟓 𝟏𝟔𝟏. 𝟔 𝒎 𝟏𝟏𝟎. 𝟔𝟔 𝒎 𝟒𝟑. 𝟑 𝒎 𝟗. 𝟒𝟔 𝒎
- 13. 𝒂 𝒔𝒊𝒏 𝑨 = 𝒃 𝒔𝒊𝒏 𝑩 = 𝒄 𝒔𝒊𝒏 𝑪 𝐎𝐁 = 𝟏𝟎𝟎 𝒔𝒊𝒏 𝟏𝟗 ° × 𝒔𝒊𝒏 𝟐𝟗 ° = 𝟏𝟒𝟖. 𝟗𝟏 𝐦 ∆ 𝑂𝐴𝐵 ∆ 𝑂𝐵𝐶 𝒄𝒐𝒔 𝟒𝟐 ° = 𝒉 𝑶𝑩 𝒉 = 𝒄𝒐𝒔 𝟒𝟐 ° × 𝟏𝟒𝟖. 𝟗𝟏 = 𝟏𝟏𝟎. 𝟔𝟔 𝒎
- 14. 𝒑𝒈 𝟐𝟐𝟔 − 𝟐𝟐𝟕
- 15. 𝒑𝒈 𝟐𝟐𝟕 − 𝟐𝟐𝟖
- 16. 𝒑𝒈 𝟐𝟑𝟎 − 𝟐𝟑𝟏
- 17. 𝒑𝒈 𝟐𝟑𝟐 𝟏𝟎 𝟑 𝟏𝟐. 𝟏 𝟒𝟏. 𝟕 𝟐𝟒 29 𝟐𝟑. 𝟒 𝟏𝟏 𝟕𝟐 ° 𝒐𝒓 𝟏𝟎𝟖 ° 𝟏𝟐𝟎 °
- 18. 𝒑𝒈 𝟐𝟑𝟐 𝟑𝟎 ° 𝟏𝟓𝟎 ° ( 𝟔 − 𝟐) 𝒄𝒎 ( 𝟔 + 𝟐) 𝒄𝒎
- 19. 𝑎 𝑏 𝑐 𝐴 𝐵 𝐶 𝑅 𝑅 𝑎 2 𝐬𝐢𝐧 𝑨 = 𝒂 𝟐 𝑹 = 𝒂 𝟐𝑹 𝒂 𝐬𝐢𝐧 𝑨 = 𝒃 𝐬𝐢𝐧 𝑩 = 𝒄 𝐬𝐢𝐧 𝑪 = 𝟐𝑹 𝒑𝒈 𝟐𝟐𝟖 − 𝟐𝟐𝟗 𝒂 𝐬𝐢𝐧 𝑨 = 𝟐𝑹 circumradius of triangle, R 三角形旳外接圆半径
- 20. 𝒑𝒈 𝟐𝟑𝟏 𝒂 𝐬𝐢𝐧 𝑨 = 𝒃 𝐬𝐢𝐧 𝑩 = 𝒄 𝐬𝐢𝐧 𝑪 = 𝟐𝑹 circumradius of triangle, R 三角形旳外接圆半径
- 21. 𝒑𝒈 𝟐𝟑𝟐 circumradius of triangle, R 三角形旳外接圆半径 𝒂 𝐬𝐢𝐧 𝑨 = 𝒃 𝐬𝐢𝐧 𝑩 = 𝒄 𝐬𝐢𝐧 𝑪 = 𝟐𝑹 𝟑𝟑. 𝟔 𝒄𝒎 𝒂 = 𝟐𝟎 𝟐 𝒄𝒎 𝒃 = 𝟐𝟎 3 𝒄𝒎 𝒄 = 𝟏𝟎 𝟔 − 𝟐 𝒄𝒎
- 22. 𝒑𝒈 𝟐𝟐𝟗 − 𝟐𝟑𝟎 𝑟 = ∆ 𝑠 inscribed circle in a triangle, r 三角形旳内切圆半径
- 23. 𝒑𝒈 𝟐𝟑𝟏 − 𝟐𝟑𝟐 𝑟 = ∆ 𝑠 = 84 21 = 4 inscribed circle in a triangle, r 三角形旳内切圆半径
Editor's Notes
- The picture on the right is a schematic diagram of the crank connecting rod structure. When the crank OA is at the left horizontal position OB, the position of the connecting rod end point P is at Q.
- Rú yòu tú, jiǎ chuán zài hé dǎo B dì zhèng nánfāng A chù, yǐ 4 hǎilǐ/xiǎoshí dì sùdù xiàng zhèng běi fāngxiàng hángxíng, qiě A dào B dì jùlí wèi 12 hǎilǐ; tóngshí, yǐ chuán yǐ 6 hǎilǐ/xiǎoshí dì sùdù qiě B dǎo chūfā, xiàng dǎo dì běi piān xī 60 degree dì fāngxiàng shǐ qù, wèn 2 xiǎoshí hòu, liǎng chuán xiāngjù duōshǎo hǎilǐ? As shown in the picture on the right, ship A is sailing at a speed of 4 nautical miles/hour to the true north at A, just south of Island B, and the distance from A to B is 12 nautical miles; meanwhile, ship B is traveling at a speed of 6 nautical miles/hour. And depart from Island B and head towards 60 degrees west of the island. After 2 hours, how many nautical miles are the two boats apart?
- https://www.ezyeducation.co.uk/ezymathsdetails/ezymaths-news/entry/countdown-to-exams-day-85-bearings.html
- In the 325 degree direction, there are three points A, B and C, AB = 7 km on the road, and one point D to the south of A, AD = 8 km, AC = 17 km, find C to D bearing, BD distance.
- One observer at the top of the cliff above the sea level h meter, the depression angles of the two boats that stop at the sea surface are 29 degrees and 48 degrees, and the line of the two boats is perpendicular to the cliff. Boat distance = 100 meter, find h. http://getdrawings.com/get-drawing#angle-of-elevation-and-depression-problems-with-solutions-and-drawing-60.jpg
- https://www.onlinemathlearning.com/herons-formula.html
- 8. m^2 + mn + n^2 = m^2 + n^2 - 2mn cos A
- 11. a=b=2, Area = 1 cm square, c=a/sin A * sin C
- https://slideplayer.com/slide/13683170/
- 12. area = 1/2 ab sin C, a = 2R sin A ... 13. L = 2 pai R, R = c/2 sin C, c = h/sin alpha