\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
W.L.O.G. let the origin be the centre of the rectangle \(ABCD\) (and let \(ABCD\) lie on the \(x\)-\(y\) plane). We adjust the scale of the axis, and we let \(V(0, 0, 1)\) and \(A(-\mu , -\nu , 0)\), we have \(B(\mu , -\nu , 0)\), \(C(\mu , \nu , 0)\) and \(D(-\mu , \nu , 0)\). Let \(\mu , \nu > 0\).
Let \(M\) be the midpoint of \(AB\) and \(N\) be the midpoint of \(BC\). We must have \(M(0, -\nu , 0)\) and \(N(\mu , 0, 0)\).
The angle between the face \(AVB\) and the base \(ABCD\) must be the angle between \(\bvect {MO}\) and \(\bvect {MV}\). Hence, \[ \cos \alpha = \frac {\bvect {MO} \cdot \bvect {MV}}{\abs *{\bvect {MO}} \abs *{\bvect {MV}}}. \]
Note that \[ \bvect {MO} = \begin {pmatrix} 0 \\ \nu \\ 0 \end {pmatrix}, \bvect {MV} = \vect {v} - \vect {m} = \begin {pmatrix} 0 \\ \nu \\ 1 \end {pmatrix}, \] and hence \[ \cos \alpha = \frac {\nu ^2}{\nu \cdot \sqrt {\nu ^2 + 1}} = \frac {\nu }{\sqrt {\nu ^2 + 1}}, \] which gives \[ \cos ^2 \alpha \nu ^2 + \cos ^2 \alpha = \nu ^2, \] and hence \[ \sin ^2 \alpha \nu ^2 = \cos ^2 \alpha , \] which gives \[ \nu = \cot \alpha . \]
Similarly, \[ \mu = \cot \beta . \]
A vector perpendicular to \(AVB\) can be \begin {align*} \bvect {VA} \times \bvect {VB} & = \begin {pmatrix} -\mu \\ -\nu \\ -1 \end {pmatrix} \times \begin {pmatrix} \mu \\ -\nu \\ -1 \end {pmatrix} \\ & = \begin {vmatrix} \ihat & \jhat & \khat \\ - \mu & -\nu & -1 \\ \mu & -\nu & -1 \end {vmatrix} \\ & = \begin {pmatrix} 0 \\ -2 \mu \\ 2\mu \nu \end {pmatrix} \\ & = \begin {pmatrix} 0 \\ -2 \cot \beta \\ 2 \cot \alpha \cot \beta . \end {pmatrix} \\ & = -\frac {2\cot \beta }{\sin \alpha } \begin {pmatrix} 0 \\ -\sin \alpha \\ \cos \alpha \end {pmatrix}, \end {align*}
and so \[ \begin {pmatrix} 0 \\ - \sin \alpha \\ \cos \alpha \end {pmatrix} \] is a unit vector perpendicular to \(AVB\).
Similarly, \begin {align*} \bvect {VB} \times \bvect {VC} & = \begin {pmatrix} \mu \\ - \nu \\ - 1 \end {pmatrix} \times \begin {pmatrix} \mu \\ \nu \\ -1 \end {pmatrix} \\ & = \begin {vmatrix} \ihat & \jhat & \khat \\ \mu & -\nu & -1 \\ \mu & \nu & -1 \end {vmatrix} \\ & = \begin {pmatrix} 2 \nu \\ 0 \\ 2 \mu \nu \end {pmatrix} \\ & = \begin {pmatrix} 2 \cot \alpha \\ 0 \\ 2 \cot \alpha \cot \beta \end {pmatrix} \\ & = \frac {2 \cot \alpha }{\sin \beta } \begin {pmatrix} \sin \beta \\ 0 \\ \cos \beta \end {pmatrix}, \end {align*}
and hence \[ \begin {pmatrix} \sin \beta \\ 0 \\ \cos \beta \end {pmatrix} \] is a unit vector perpendicular to \(BVC\).
The acute angle between \(AVB\) and \(BVC\), \(\theta \), satisfies that \[ \cos \theta = \begin {pmatrix} 0 \\ - \sin \alpha \\ \cos \alpha \end {pmatrix} \cdot \begin {pmatrix} \sin \beta \\ 0 \\ \cos \beta \end {pmatrix} = \cos \alpha \cos \beta , \] as desired.
Notice that \begin {align*} \cos \phi & = \frac {\bvect {BV} \cdot \bvect {BO}}{\abs *{\bvect {BV}} \cdot \abs *{\bvect {BO}}} \\ & = \frac {\begin {pmatrix} -\mu \\ \nu \\ 1 \end {pmatrix} \cdot \begin {pmatrix} -\mu \\ \nu \\ 0 \end {pmatrix}}{\sqrt {\mu ^2 + \nu ^2 + 1} \sqrt {\mu ^2 + \nu ^2}} \\ & = \sqrt {\frac {\mu ^2 + \nu ^2}{\mu ^2 + \nu ^2 + 1}}, \end {align*}
and hence \[ \sin \phi = \sqrt {1 - \cos ^2 \phi } = \sqrt {\frac {1}{\mu ^2 + \nu ^2 + 1}}, \] which means \[ \cot \phi = \sqrt {\mu ^2 + \nu ^2}, \] and hence \[ \cot ^2 \phi = \mu ^2 + \nu ^2 = \cot ^2 \alpha + \cot ^2 \beta , \] as desired.
Notice that \begin {align*} \cos ^2 \phi & = \frac {\mu ^2 + \nu ^2}{\mu ^2 + \nu ^2 + 1} \\ & = \frac {\cot ^2 \alpha + \cot ^2 \beta }{\cot ^2 \alpha + \cot ^2 \beta + 1} \\ & = \frac {\cos ^2 \alpha \sin ^2 \beta + \cos ^2 \beta \sin ^2 \alpha }{\cos ^2 \alpha \sin ^2 \beta + \cos ^2 \beta \sin ^2 \alpha + \sin ^2 \beta \sin ^2 \alpha } \\ & = \frac {\cos ^2 \alpha (1 - \cos ^2 \beta ) + \cos ^2 \beta (1 - \cos ^2 \alpha )}{(\cos ^2 \alpha + \sin ^2 \alpha )(\cos ^2 \beta + \sin ^2 \beta ) - \cos ^2 \alpha \cos ^2 \beta } \\ & = \frac {\cos ^2 \alpha + \cos ^2 \beta - 2 \cos ^2\alpha \cos ^2\beta }{1 - \cos ^2 \alpha \cos ^2 \beta } \\ & = \frac {\cos ^2 \alpha + \cos ^2 \beta - 2 \cos ^2\theta }{1 - \cos ^2\theta }. \end {align*}
Since \((\cos \alpha - \cos \beta )^2 = \cos ^2 \alpha + \cos ^2 \beta - 2 \cos \theta \geq 0\), we have \(\cos ^2 \alpha + \cos ^2 \beta \geq 2 \cos \theta \), and hence \[ \cos ^2 \phi = \frac {\cos ^2 \alpha + \cos ^2 \beta - 2 \cos ^2\theta }{1 - \cos ^2\theta } \geq \frac {2 \cos \theta - 2 \cos ^2 \theta }{1 - \cos ^2 \theta }. \]
Notice that \begin {align*} \cos ^2 \phi & \geq \frac {2 \cos \theta - 2 \cos ^2 \theta }{1 - \cos ^2 \theta } \\ & = \frac {2 \cos \theta (1 - \cos \theta )}{(1 - \cos \theta )(1 + \cos \theta )} \\ & = \frac {2 \cos \theta }{1 + \cos \theta } \\ & = \frac {2}{1 + \cos \theta } \cos \theta \\ & > \frac {2}{1 + 1} \cos \theta \\ & = \cos \theta \\ & > \cos ^2 \theta , \end {align*}
since \(\theta \) is acute, \(0 < \cos \theta < 1\).
This means \(\cos ^2 \phi > \cos ^2 \theta \), and since \(\theta , \phi \) are acute, this must mean that \(\phi < \theta \), since \(\cos \phi , \cos \theta \) are both positive, and \(\cos \phi > \cos \theta \).