✔ 最佳答案
設H點係AB上,AH = BH = 10cm
設K點係AD上,AK = DK = 10.5 cm
BD = √[(20^2) + (21^2)] (畢氏定理)
BD = 29 cm
VH = √[(20^2) + (10.5^2)] (畢氏定理)
VH = 22.588714 cm
tan∠VAB = VH / AH
∠VAB = 66.12107809°
VK = √[(20^2) + (10^2)] (畢氏定理)
VK = 22.36067978 cm
tan∠VAD = VK / AK
∠VAD = 64.84645763°
設平面VAB與平面VAD的交角是∠DZB
DZ = 21sin∠VAD
DZ = 19.00861186 cm
BZ = 20sin∠VAB
BZ = 18.28805875 cm
在△DZB,
cos∠DZB = [(DZ)^2 + (BZ)^2 - (BD)^2] / 2(DZ)(BZ)
∠DZB = 102.0561709°
因此,平面VAB與平面VAD的交角是102.°(準確至3位有效數字)