請詳細解釋下條做法原因 :
不要網址回答 :
圖片參考：http://imgcld.yimg.com/8/n/HA05788109/o/20130126142327.jpg
(a) 有人教我做法如下,但我唔明條式點來,又唔明點解要咁做.
cosB + cosD
= 2cos[(B + D)/2]cos[(B - D)/2]
= 2cos90°cos[(B - D)/2]
= 2(0)cos[(B - D)/2]
= 0
急! F5 Trigonometry 1題11q3a
2013-01-26 10:25 pm
回答 (2)
2013-01-27 4:18 am
✔ 最佳答案
26.(a)
B + D = 180°
Then, B = 180° - D
L.H.S.
= cosB + cosD
= cos(180° - D) + cosD
= -cosD + cosD
= 0
= R.H.S.
Hence, cosB + cosD = 0
(b)
In ΔABC :
cosB = (AB² + BC² - AC²) / (2 x AB x BC) (cosine law)
cosB = (4² + 4² - AC²) / (2 x 4 x 4)
cosB = (32 - AC²) / 32
In ΔADC :
cosD = (DA² + CD² - AC²) / (2 x DA x CD) (cosine law)
cosD = (3² + 6² - AC²) / (2 x 3 x 6)
cosD = (45 - AC²) / 36
In (a), it has been proven that :
cosB + cosD = 0
[(32 - AC²) / 32] + [(45 - AC²) / 36] = 0
(32 - AC²) / 32 = -(45 - AC²) / 36
(32 - AC²) x 36 = -(45 - AC²) x 32
1152 - 36 AC² = -1440 + 32 AC²
68 AC² = 2592
AC² = 2592/68
AC = 6.17 (cm)
參考: 土扁
2013-01-26 10:59 pm
It is the Sum-to-product formulae only.
收錄日期: 2021-04-13 19:16:10
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