求三角學 20點 - Ans.fyi 參考答案

求三角學 20點

用戶22760

2011-11-19 4:08 am

1. 圖中，ABCD為四邊形。 ∠ABC=90°，∠ACB=40°，∠ACD=80°，
∠ADC=50°及AD=27cm。
(a) 求AC的長度
(b) 求AB的長度

圖片參考：http://imgcld.yimg.com/8/n/HA00080227/o/701111180071613873403030.jpg

2. 一船從A處測得燈塔B在N27°E，若該船由A向S54°E行駛20km抵達C地，再測燈塔B，得方位為N8°W，求這時燈塔和船的距離。答案準確至兩位小數。

圖片參考：http://imgcld.yimg.com/8/n/HA00080227/o/701111180071613873403031.jpg

3. 圖中，ABCD為一個平行四邊形。AD=10cm，CD=14cm及∠ADC=36°
(a) 求AC的長度
(b) 求∠BCD
(c) 求BD的長度

圖片參考：http://imgcld.yimg.com/8/n/HA00080227/o/701111180071613873403042.jpg

4. A、B及C是水平地面上的三點，P位於C的正上方。已知AB=10km，∠CAB=40°及∠CBA=20°。
(a) 求AC和BC的長度，準確至最接近0.1km。
(b) 若由B測得P的仰角是30°，求
(i) 由B測得P的仰角;
(ii) PA與PB的交角，準確至最接近的度。

回答 (1)

用戶28092

2011-11-25 1:50 am

✔ 最佳答案

1a.
by sine law at triangle ADC:
AC/sin 50 = 27/sin 80 => AC = ...
1b.
by sine law at triangle ABC:
AC/sin 90 = AB/sin 40 => AB = ...

2.
angle ABC = 35
angle BAC = 99
20/sin 35 = BC/sin 99 => BC = ...

3a.
by cosine law at triangle ADC
AC^2 = 10^2 + 14^2 - 2(10)(14)cos 36 => AC = ...
3b.
angle ADC + angle BCD = 180 => angle BCD = ...
3c.
by cosine law at traingle BCD
BD^2 = 10^2 + 14^2 - 2(10)(14)cos (angle BCD) => BD = ...

4a.
by sine law at triangle ABC
10/sin 40 = AC/sin 20 = BC/sin (180-40-20) => AC, BC = ...
4bi.
suppose the question is asking the elevation from A.
by sine law at triangle BPC
PC/sin 30 = BC/sin 60 => PC = ...
at triangle APC
tan (angle PAC) = PC/AC => angle PAC = ...
4bii.
at triangle PAC
PA^2 = PC^2 + AC^2 => PA = ...
at triangle PAB
PB^2 = PC^2 + BC^2 => PB = ...
by cosine law at triangle PAB
AB^2 = PA^2 + PB^2 - 2(PA)(PB)cos (angle APB) => angle APB = ...

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