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SUEC 高中 Adv Maths (Sin and Cos Rule)
- 1. 𝒑𝒈 𝟐𝟏𝟎 − 𝟐𝟏𝟐 𝒂𝒏𝒈𝒍𝒆 𝒍𝒆𝒏𝒈𝒕𝒉 𝒍𝒆𝒏𝒈𝒕𝒉
- 3. 𝒑𝒈 𝟐𝟏𝟎 − 𝟐𝟏𝟐 𝒂𝒏𝒈𝒍𝒆 𝒍𝒆𝒏𝒈𝒕𝒉
- 4. 𝒑𝒈 𝟐𝟏𝟎 − 𝟐𝟏𝟑 𝒂𝒏𝒈𝒍𝒆 ?
- 5. 𝒑𝒈 𝟐𝟏𝟒 𝐬𝐢𝐧 𝐁 = 𝒔𝒊𝒏 𝟏𝟓𝟎 ° 𝟏𝟖 × 𝟐𝟎 = 𝟓 𝟗 𝐁 = 𝟑𝟑. 𝟕𝟓 𝐨𝐫 𝟏𝟒𝟔. 𝟐𝟓 ∴ 𝐧𝐨 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐚𝐬 𝐭𝐡𝐢𝐬 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞 𝐝𝐨𝐞𝐬 𝐧𝐨𝐭 𝐞𝐱𝐢𝐬𝐭.
- 6. 𝒑𝒈 𝟐𝟏𝟔 𝒂𝒏𝒈𝒍𝒆 𝒍𝒆𝒏𝒈𝒕𝒉 𝑩 = 𝟑𝟓 ° 𝟏𝟔′ 𝐜 = 𝟏𝟕. 𝟐𝟓 𝑪 = 𝟖𝟒 ° 𝟒𝟒′ 𝒔𝒊𝒏 𝑨 = 𝟐𝟓 𝟐𝟐 > 𝟏 𝒏𝒐 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑨 = 𝟗𝟕 ° 𝟏𝟖′ 𝑩 = 𝟓𝟕 ° 𝟒𝟐′ 𝐚 = 𝟒𝟔. 𝟗𝟒 𝑨 = 𝟑𝟐 ° 𝟒𝟐′ 𝑩 = 𝟏𝟐𝟐 ° 𝟏𝟖′ 𝐚 = 𝟐𝟓. 𝟓𝟕 𝒔𝒊𝒏 𝑩 = 𝟏. 𝟑𝟔 > 𝟎 𝒏𝒐 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝑨 = 𝟏𝟒 ° 𝟐𝟗′ 𝑩 = 𝟔𝟐 ° 𝟕′ 𝐜 = 𝟏𝟏. 𝟎𝟏 𝑨 = 𝟏𝟒 ° 𝟐𝟗′ 𝑩 = 𝟏𝟏𝟕 ° 𝟓𝟑′ 𝐜 = 𝟖. 𝟑𝟔 𝑪 = 𝟏𝟎𝟑 ° 𝟐𝟒′ 𝑪 = 𝟒𝟕 ° 𝟑𝟖′ 𝑩 = 𝟒𝟓 ° 𝐜 = 𝟑. 𝟖𝟔 cm 𝑩 = 𝟏𝟑𝟓 ° 𝐜 = 𝟏. 𝟎𝟔 𝐜𝐦 𝑪 = 𝟏𝟎𝟓 ° 𝟕′ 𝑪 = 𝟏𝟓 °
- 7. 𝒑𝒈 𝟐𝟏𝟔 𝒂𝒏𝒈𝒍𝒆 𝒍𝒆𝒏𝒈𝒕𝒉 𝑨 = 𝟕𝟎 ° 𝟏𝟐′ 𝑩 = 𝟓𝟕 ° 𝟐𝟒′ 𝐛 = 𝟐𝟖. 𝟕𝟗 𝐜𝐦 𝑨 = 𝟏𝟎𝟗 ° 𝟒𝟖′ 𝑩 = 𝟏𝟕 ° 𝟒𝟖′ 𝐚 = 𝟕. 𝟓𝟑 𝐜𝐦 𝐛 = 𝟓. 𝟏𝟔 𝐜𝐦 𝑪 = 𝟖𝟑 ° 𝟑𝟗′ 𝐛 = 𝟏𝟎. 𝟒𝟓 𝐜𝐦 两边为 𝟓𝟎. 𝟔𝟑 𝒄𝒎, 𝟔𝟎. 𝟒𝟒 𝒄𝒎 𝑨𝒓𝒆𝒂 = 𝟑𝟎𝟓𝟒 𝒄𝒎𝟐 𝐚 = 𝟕. 𝟑𝟗 𝐜𝐦 𝐛 = 𝟏𝟑. 𝟖𝟗 𝐜𝐦 𝟖𝟏. 𝟒𝟔 cm 𝐜 = 𝟏𝟖. 𝟕𝟐 𝐜𝐦
- 8. 𝒑𝒈 𝟐𝟏𝟔 − 𝟐𝟏𝟖 𝒍𝒆𝒏𝒈𝒕𝒉 𝒂𝒏𝒈𝒍𝒆 b
- 9. 𝒑𝒈 𝟐𝟐𝟏 𝒍𝒆𝒏𝒈𝒕𝒉 𝒂𝒏𝒈𝒍𝒆 𝟗𝟎 ° 𝟕 𝒄𝒎 𝟗𝟖 ° 𝟏𝟑′
- 11. 𝒑𝒈 𝟐𝟏𝟗 − 𝟐𝟐𝟎
- 12. 𝒑𝒈 𝟐𝟐𝟏 𝒍𝒆𝒏𝒈𝒕𝒉 𝒂𝒏𝒈𝒍𝒆 𝑨 = 𝟐𝟏 ° 𝟒𝟕′ 𝑩 = 𝟖 ° 𝟏𝟑′ 𝑪 = 𝟏𝟓𝟎 ° 𝑨 = 𝟓𝟗 ° 𝟐𝟗′ 𝑩 = 𝟗𝟎 ° 𝑪 = 𝟑𝟎 ° 𝟑𝟏′ 𝑨 = 𝟑𝟔 ° 𝟐𝟑′ 𝒃 = 𝟔𝟖. 𝟖𝟓 𝑪 = 𝟓𝟖 ° 𝟑𝟕′ 𝒂 = 𝟔𝟐. 𝟑𝟏 𝑩 = 𝟑𝟓 ° 𝟓𝟒′ 𝑪 = 𝟑𝟖 ° 𝟔′ 𝑨 = 𝟒𝟓 ° 𝑩 = 𝟑𝟎 ° 𝑪 = 𝟏𝟎𝟓 ° 𝑨 = 𝟒𝟒 ° 𝟒𝟐′ 𝑩 = 𝟓𝟕 ° 𝟑𝟔′ 𝒄 = 𝟕𝟕. 𝟏𝟓 cm 𝟏𝟎𝟒 ° 𝟐𝟗′ 𝟐. 𝟎𝟕 𝒙 𝟒𝟏 ° 𝟑𝟕′ 𝟑𝟎 °
- 13. 𝒑𝒈 𝟐𝟐𝟏 𝒍𝒆𝒏𝒈𝒕𝒉 𝒂𝒏𝒈𝒍𝒆 𝑐𝑜𝑠 𝐴 = 𝑏2 + 𝑐2 − 𝑎2 2𝑏𝑐 = 2 2 + 3 + 1 2 − 22 2 × 2 × 3 + 1 = 2 + 3 + 2 3 + 1 − 4 2 2 3 + 1 = 2 + 2 3 2 2 3 + 1 = 2 1 + 3 2 2 3 + 1 𝐴 = 45 ° 𝑠𝑖𝑛 𝐵 𝑏 = 𝑠𝑖𝑛 𝐴 𝑎 𝑠𝑖𝑛 𝐵 = 𝑠𝑖𝑛 𝐴 𝑎 × 𝑏 = 𝑠𝑖𝑛 45 ° 2 × 2 𝐵 = 30 ° 𝐶 = 180 ° − (𝐴 + 𝐵) = 180 ° − (45 ° + 30 °) 𝐶 = 105 °
- 14. 𝒑𝒈 𝟐𝟐𝟏 𝒍𝒆𝒏𝒈𝒕𝒉 𝒂𝒏𝒈𝒍𝒆 𝑠𝑖𝑛 𝐴 𝑎 = 𝑠𝑖𝑛 𝐵 𝑏 𝑠𝑖𝑛 𝐴 = 𝑠𝑖𝑛 𝐵 𝑏 × 𝑎 = 𝑠𝑖𝑛 95 ° 3𝑥 × 2𝑥 𝐴 = 41 ° 37’ = 0.6641 𝐶 = 180 ° − (𝐴 + 𝐵) = 180 ° − (41 °37’ + 95 °) = 43.38 ° 𝑐2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 𝑐𝑜𝑠 𝐶 = (2)2 +(3𝑥)2 − 2 2𝑥 3𝑥 𝑐𝑜𝑠 43.38 ° = 13𝑥2 − 8.72𝑥2 c = 2.07 𝑥
- 15. 𝒑𝒈 𝟐𝟐𝟏 𝒍𝒆𝒏𝒈𝒕𝒉 𝒂𝒏𝒈𝒍𝒆 𝒂: 𝒃: 𝒄 = 𝟕 ∶ 𝟔. 𝟗𝟑 ∶ 𝟑. 𝟔𝟏 𝑐𝑜𝑠 𝐶 = 𝑎2 + 𝑏2 − 𝑐2 2𝑎𝑏 = 72 + (4 3)2−( 13)2 2 × 7 × 4 3 𝐶 = 30 °
- 16. 𝒑𝒈 𝟐𝟐𝟏 𝒍𝒆𝒏𝒈𝒕𝒉 𝒂𝒏𝒈𝒍𝒆 𝟏𝟎𝟏. 𝟕𝟑 𝒄𝒎 𝟖𝟐 ° 𝟒𝟗′
- 17. 𝒑𝒈 𝟐𝟐𝟏 𝒍𝒆𝒏𝒈𝒕𝒉 𝒂𝒏𝒈𝒍𝒆 𝟐 𝟑 𝟐. 𝟓 ? 𝟐. 𝟓 ∆ 𝑨𝑩𝑪 𝑐𝑜𝑠 𝐵 = 𝑎2 + 𝑐2 − 𝑏2 2𝑎𝑐 = (2.5)2 +22 − 32 2 × 2.5 × 2 𝑎𝑛𝑔𝑙𝑒 𝐶𝐵𝐴 = 82 ° 49’
- 18. 𝒑𝒈 𝟐𝟐𝟏 𝒍𝒆𝒏𝒈𝒕𝒉 𝒂𝒏𝒈𝒍𝒆 𝟑𝟗. 𝟑𝟐 𝒄𝒎 𝟐𝟖. 𝟎𝟎 𝒄𝒎 𝒂 𝒃 𝒄 𝑩𝑸𝟐 = 𝑨𝑸𝟐 + 𝒄𝟐 − 𝟐𝒂𝒄 𝒄𝒐𝒔 𝑨 = (𝟑𝟗. 𝟑𝟐)𝟐 +(𝟐𝟒. 𝟒𝟗)𝟐 − 𝟐 𝟑𝟗. 𝟑𝟐 𝟐𝟒. 𝟒𝟗 𝒄𝒐𝒔 𝟒𝟓 ° 𝒂 𝒔𝒊𝒏 𝑨 = 𝒃 𝒔𝒊𝒏 𝑩 = 𝒄 𝒔𝒊𝒏 𝑪 𝐛 = 𝟐𝟎 𝒔𝒊𝒏 𝟒𝟓 ° × 𝒔𝒊𝒏 (𝟏𝟖𝟎 − 𝟏𝟎𝟓) ° = 𝟐𝟕. 𝟑𝟐 𝐜𝐦 𝐀𝐐 = 𝟐𝟕. 𝟑𝟐 + 𝟏𝟐 = 𝟑𝟗. 𝟑𝟐 𝐜𝐦 ∆ 𝑨𝑩𝑪 𝐜 = 𝟐𝟎 𝒔𝒊𝒏 𝟒𝟓 ° × 𝒔𝒊𝒏 (𝟏𝟎𝟓 − 𝟒𝟓) ° = 𝟐𝟒. 𝟒𝟗 𝐜𝐦 ∆ 𝑨𝑩𝑸 𝑩𝑸 = 𝟐𝟖. 𝟎𝟎 𝒄𝒎
Editor's Notes
- https://www.onlinemathlearning.com/triangle-questions.html
- https://www.onlinemathlearning.com/triangle-questions.html
- https://www.onlinemathlearning.com/triangle-questions.html
- 3. one diagonal of the parallelogram is 76.33 cm long, and this diagonal forms an angle A and B with the two sides respectively.
- https://www.onlinemathlearning.com/cosine-rule.html
- 5. A = 28.96 degree, B = 180 – A = 151.04 degree