數學問題一條(立體)急...

2010-02-06 4:45 am

下圖顯示有一幢水平底ABC的大廈,其中AB=BC=CA=10m.P,Q,R分別埀直於A,B,C的上方使PA=30m,QB=20m和RC=10m.
(a)證PQR為等腰三角形
(b)求tan角APR和tan角APQ
(c)M和N分別為PR和PA上的一點使QM和NM與PR埀直
(i)求QM.MN,QN.答案以根式或準確值表示
(ii)由此證平面PQR和APRC是埀直對方

圖片參考：http://imgcld.yimg.com/8/n/HA00543461/o/701002050131113873397470.jpg

更新1:

c對上就係R

更新2:

有無人可以幫下我!!

回答 (1)

用戶2053

2010-02-06 9:45 am

✔ 最佳答案

a)PQ^2 = (PA-QB)^2 + AB^2 = (30-20)^2 + 10^2 = 200
QR^2 = (QB - RC)^2 + BC^2 = (20-10)^2 + 10^2 = 200
So PQ^2 = QR^2
PQ = QR = 10√2
PQR為等腰三角形
b)tan ㄥAPR = AC / (PA - RC) = 10 / (30 - 10) = 1/2
tan ㄥAPQ = AB / (PA - QB) = 10 / (30-20) = 1
(即ㄥAPQ = 45°)
c)i) PR^2 = (PA-RC)^2 + AC^2 = (30-10)^2 + 10^2 = 500
PR = 10√5
PM = PR/2 = 5√5
QM^2 = PQ^2 - PM^2 = 200 - 125 = 75
QM = √75 = 5√3 m
MN/PM = tanㄥAPR
MN/(5√5) = 1/2
MN = 5√5/2 m
PN^2 = PM^2 + MN^2 = (5√5)^2 + (5√5/2)^2 = 125 + 125/4 = 625/4
由餘弦定理 :
QN^2 = PN^2 + PQ^2 - 2(PN)(PQ)cosㄥAPQ
QN^2 = 625/4 + 200 - 2(25/2)(10√2) cos 45°
QN^2 = 1425/4 - 25(10√2)(√2/2) = 425/4
QN = √(425/4) = 5√17/2 m
C)ii) QN^2 = 425/4
QM^2 = 75
MN^2 = (5√5/2)^2 = 125/4
425/4 = 75 + 125/4
QN^2 = QM^2 + MN^2
ㄥQMN是直角,即
平面PQR和APRC是埀直對方。

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