In the figure, VABCD is a right pyramid with a square base ABCD of side
10 cm. The length of each slant edge is 15 cm. E and F are the mid - points
of AB and BC respectively. Find
a) VE and EF,
b) the angle between VE and EF,
c) the angle between the planes VEF and ABCD.
Figure: http://img511.imageshack.us/img511/9771/maths39a.jpg
F4 Maths 3-D problems @\”@
2010-03-24 6:09 am
回答 (2)
2010-03-24 7:24 am
✔ 最佳答案
a 5^2+VE^2=15^2VE = 開方200
EF^2=EB^2+BF^2
EF^2=5^2+5^2
EF = 開方50
b COS<VEC= (開方200)^2+(開方50)^2-(開方200)^2 [OVER] 2*開方200*開方50
COS<VEC=0.25
<VEC=75.522488 (我唔知你要幾多個位...)
參考: 用手計~_~
2010-03-24 7:09 am
a) AE=BE=5cm
VE^2 = 15^2-5^2
VE=square root 200
BE=BF=5cm
EF^2 = 5^2+5^2
EF= square root 50
b) 90o
because VE is a vertical line and EF is a horizontal line.
(angle VEF)
c)let R be the mid point of EF
therefore the angle is angle VRD
VE^2 = 15^2-5^2
VE=square root 200
BE=BF=5cm
EF^2 = 5^2+5^2
EF= square root 50
b) 90o
because VE is a vertical line and EF is a horizontal line.
(angle VEF)
c)let R be the mid point of EF
therefore the angle is angle VRD
收錄日期: 2021-04-29 21:37:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100323000051KK01723