In Figure 9(a), ABC is a triangular paper card. D is a point lying on AC such that BD is perpendicular to AC. It is known that AB = 40 cm, BC = 60 cm and AC = 90 cm.
圖片參考：http://img176.imageshack.us/img176/8537/53559974gn4.jpg
http://img176.imageshack.us/img176/8537/53559974gn4.jpg
(a) Find AD.
(b) The triangular card in Figure 9(a) is folded along BD such that AB and BC lie on a horizontal plane as shown in Figure 9(b).
圖片參考：http://img352.imageshack.us/img352/4041/51771486zq2.jpg
http://img352.imageshack.us/img352/4041/51771486zq2.jpg
(i) Suppose ∠DAC = 62
(1) Find the distance between A and C on the horizontal plane.
(2) Using Heron's formula, or otherwise, find the area of △ABC on the horizontal plane.
(3) Find the height of the tetrahedron ABCD from the vertex D to the base △ABC.
(ii) Describe how the volume of the tetrahedron ABCD varies when ∠ADC increases from 30 to 150. Explain your answer.
Trigonometry (3)
2008-08-24 6:03 am
回答 (2)
2008-08-24 10:47 am
✔ 最佳答案
(a) Find AD圖片參考：http://hk.geocities.com/namen1997/p0031A.jpg
(b) The triangular card in Figure 9(a) is folded along BD such that AB and BC lie on a horizontal plane as shown in Figure 9(b).
(i) Suppose ∠DAC = 62
圖片參考：http://hk.geocities.com/namen1997/p0031B.jpg
圖片參考：http://hk.geocities.com/namen1997/p0031C.jpg
(ii) Describe how the volume of the tetrahedron ABCD varies when ∠ADC increases from 30 to 150. Explain your answer.
圖片參考：http://hk.geocities.com/namen1997/p0031D.jpg
圖片參考：http://hk.geocities.com/namen1997/p0031E.jpg
2008-08-24 8:36 am
AD ?
Using cosine law
let angle bac be y
cos y = [40^2+90^2-60^2]/2(40)(90)
y = 32.08918386
in triangle abd cos y = ad / 40, ad= 33.8888888889
then , we use cosine law to solve the other question
∠DAC = 62
cos 62 = 33.8888888889^2+ac^2-(90-33.8888888889)^2/(2*33.8888888889ac)
you can solve it itself!
Using Heron's formula, or otherwise, find the area of △ABC on the horizontal plane.
just too easy!
s =( the sum of 3 side )/2
area = rt [s*s-(ab)*s-ac*s-bc]
Using cosine law
let angle bac be y
cos y = [40^2+90^2-60^2]/2(40)(90)
y = 32.08918386
in triangle abd cos y = ad / 40, ad= 33.8888888889
then , we use cosine law to solve the other question
∠DAC = 62
cos 62 = 33.8888888889^2+ac^2-(90-33.8888888889)^2/(2*33.8888888889ac)
you can solve it itself!
Using Heron's formula, or otherwise, find the area of △ABC on the horizontal plane.
just too easy!
s =( the sum of 3 side )/2
area = rt [s*s-(ab)*s-ac*s-bc]
參考: if i do all for you, you learn nothing
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