amaths~~~~~~

VABC is a right pyramid whose base ABC is an equilateral triangle . AB = 12cm and VA = 24cm . D is a point on VB such that AD is perpendicular to VB . Find , correct to 3 sig.fig ,

圖片參考：http://hk.geocities.com/namsm4e/p49.jpg

(a) ∠VBA and AD
cos∠VBA = 0.5AB / AV = 0.5x12 / 24 = 0.25
∠VBA = 75.5o
AD = AB sin ∠VAB = 12 sin 75.5 o
AD = 11.6cm

(b) the angle between the faces VAB and VBC
AD⊥VB 而 CD⊥VB
及 CD = AD = 11.6
所以兩面的夾角為θ
用餘弦定律
AC2 = AD2 + CD2 – 2ADxCD cosθ
122 = 11.62 + 11.62 – 2x11.6x11.6 cosθ
cosθ = -126 / 270
θ = 118o

Let E be the mid-point of AB such that AE = BE = 6cm.
In triangle AEV,
cos angle VAE = AE/AV
cos angle VAE = 6/24 = 1/4
angle VAE = 75.52248781 degrees
angle VAB = 75.52248781 degrees
area of triangle VAB
= 2 x area of triangle VAE = 2 x 1/2 x VE x AE
= 1/2 x VB x AD
VE = √(24²-6²) [pyth. th.] = √540 = 6√15
2 x 1/2 x 6√15 x 6 = 1/2 x 24 x AD
36√15 = 12 AD
AD = 3√15 cm
AD = CD = 3√15 cm
AC = 12 cm
in triangle ACD, by cosine formula,
cos angle ACD = (AD²+CD²-AC²)/2(AD)(CD)
cos angle ACD = [(3√15)²+(3√15)²-12²)/2(3√15)( 3√15)
angle ACD = 62.18186072 degrees
the angle between the faces VAB and VBC is 62.18 degrees.

2007-03-23 22:12:30 補充：
sorry, angle VAB should be angle VBA...Let E be the mid-point of AB such that AE = BE = 6cm.In triangle BEV,cos angle VBE = BE/BVcos angle VBE = 6/24 = 1/4angle VBE = 75.52248781 degreesangle VBA = 75.52248781 degrees