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立体几何命题中,求二面角的值是一种常见而且重要的问题。一般的做法是先找出二面角的平面角再计算。本文拟给出一个直接求二面角的公式,并讨论一些相关问题。 定理 设二面角M-AB-N的大小为a,P∈AB,D∈平面N,C∈平面M,∠CPB=θ_1,∠DPB=θ_2,∠CPD=θ,则有 cosθ-cosθ_1cosθ_2 证明:如图1,作AB的垂面,分别交PC、AB、PD于C、E、D.则∠CED=a,∠CEP=∠DEP=90°.设PE=x,从而有PC=xsecθ_1,EC=xtgθ_1,PD=xsecθ_2,DE=xtgθ_2. 在△PCD与△ECD中,分别用余弦定理求CD~2,得整理得 应用此定理便可直接求出二面角的值,请看下面的例子。
In the three-dimensional geometry proposition, the value of the dihedral angle is a common and important problem. The general approach is to first find out the dihedral plane angle and calculate it. This paper proposes a formula for direct dihedral angles and discusses some related issues. Theorem sets the size of the dihedral angle M-AB-N as a, P∈AB, D∈plane N, C∈plane M, ∠CPB=θ_1, ∠DPB=θ_2, ∠CPD=θ, then there is cosθ-cosθ_1cosθ_2 : As shown in Figure 1, for the AB vertical surface, respectively, pay PC, AB, PD in C, E, D. Then ∠ CED = a, ∠ CEP = ∠ DEP = 90 °. Let PE = x, so there is PC = xsecθ_1 , EC = xtgθ_1, PD = xsecθ_2, DE = xtgθ_2. In △ PCD and △ ECD, respectively, using cosine theorem to find CD ~ 2, have to sort out the application of this theorem can be directly found dihedral value, see below example of.