畢氏定理,幫幫手,我唔識

[匿名]

2011-04-18 9:44 am

圖片參考：http://imgcld.yimg.com/8/n/HA06959049/o/701104180008513873428050.jpg

1.如圖所示,晾衣架AB的長度為2.4m,它由一條垃緊的繩索AC緊擊至牆壁上。若B與C相距0.7m,問繩索AC的長度是多少??

圖片參考：http://imgcld.yimg.com/8/n/HA06959049/o/701104180008513873428051.jpg

2.如圖所示,一把長2.6m的梯子AB斜擱在一堵直立的牆壁上,梯腳與牆壁相距1m。現把梯子的頂端沿牆壁下移0.4m,問梯腳向外移了多少?(答案須準確至二位小數。)

圖片參考：http://imgcld.yimg.com/8/n/HA06959049/o/701104180008513873428062.jpg

3.思郎騎單車從位置A向西行駛了4km,再向北行駛了3km到達位置B。
a)求A與B之間的距離。
b)思郎到達位置B後,再轉向西行駛了5km到達位置C。求A與C之間的距離。

圖片參考：http://imgcld.yimg.com/8/n/HA06959049/o/701104180008513873428063.jpg

4. 圖中所示的天線杆由兩條鋼索來固定位置。已知每條鋼索均緊繫着天線杆離地面50cm處與相距杆底25cm的地面上的一點,求該兩條鋼索的總長度。

回答 (2)

Godfrey

2011-04-18 11:31 am

✔ 最佳答案

1.
AC^2 = AB^2 + BC^2
AC^2 = (2.4 m)^2 + (0.7 m)^2
AC^2 = (2.4 m)^2 + (0.7 m)^2 = 6.25 m^2
AC = (6.25 m^2)^0.5
AC = 2.5 m

2.
AB^2 = BC^2 + AC^2
(2.6 m)^2 = (1 m)^2 + AC^2
AC^2 = (2.6 m)^2 – 1 m^2 = 5.76 m^2
AC = (5.76 m)^0.5 = 2.4 m
The ladder slided to new position A to A’, B to B’
AC’ = AC – AA” = 2.4 m – 0.4 m = 2 m
A’B’ = A’C^2 + B’C^2
(2.6 m)^2 = (2 m)^2 + B’C^2
B’C^2 = (2.6 m)^2 - (2 m)^2 = 2.76 m^2
B’C = 1.66 m
B’B = B’C – BC = 1.66 m – 1 m = 0.66 m
梯腳向外移了 0.66 m

3.
AB^2 = BE^2 + EA^2
AB^2 = ( 3 km)^2 + (4 km) = 25 km^2
AB = (25 km)^0.5 = 5 km
AB = 5 km

Tan angle ABE = EA/BE = 4/3
angle ABE = 53.13°
angle CBA = 90° + 53.13° = 143.13°
Using Cosine formula
AC^2 = BC^2 + AB^2 - 2(BC)(AB) cos angle CBA
AC^2 = (5 km)^2 + (5 km)^2 - 2(5 km)(5 km) cos 143.13°
AC^2 = 50 km^2 + 50 km (0.8) = 90 km^2
AC = (90 km^2)^0.5 = 9.49 km
AC = 9.49 km

4.
(一條鋼索長度)^2 = (50 cm)^2 + (25 cm)^2 = 3125 cm^2
一條鋼索長度 = (3125)^0.5 = 55.90 cm
兩條鋼索總長度 = 55.90 cm x 2 =111.80 cm

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一個青檸

2011-04-18 10:10 pm

not clear ...

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