I am attempting to find the height of a v-fold for a popup book.
I've been pondering this problem, and I think the best way to abstract it is by imagining that I'm actually dealing with a full parrallelpiped (instead of the half one that actually exists) and then attempt to find the height of it.
In my research I have found one formula (here) for calculating the height:
h = ∥c∥|cosϕ|
Where c is the vector of the "vertical" side and ϕ is the angle between the vector c and the vertical. 
That all makes sense, but it's not very helpful since I don't know the angle ϕ And I don't have enough experience to know how to calculate it.
As for my level of knowledge: I am an undergraduate student, and I have not yet taken a course in 3D Geometry. I would love to see theorems (since I'm quite interested in math). But I would also really appreciate a formula/algorithm for calculating the height given the side lengths and angles between them, if that is possible.
Thank you for your time! --Beka
If you know only $c$ and the angles $\alpha = \angle(\vec{a}, \vec{c}), \, \beta = \angle(\vec{b}, \vec{c})$ and $\gamma = \angle(\vec{a}, \vec{b})$, then you can calculate the height $h$ from the tip of $\vec{c}$ to the plane spanned by $\vec{a}$ and $\vec{b}$. You do not need the magnitudes of the latter two vectors. The formula is $$h = c \, \frac{\sqrt{\,1 + 2\, \cos(\alpha) \cos(\beta) \cos(\gamma) - \cos^2(\alpha) - \cos^2(\beta) - \cos^2(\gamma)\,}}{\sin(\gamma)}$$
A geometric solution. I will follow the notations from the picture. $A$ and $B$ are the orthogonal projections of the segment $OC$ onto the vectors $\vec{a}$ and $\vec{b}$ respectively. Therefore $$\angle \, OAC = \angle \, OBC = 90^{\circ}$$ and thus the two triangles $\Delta \, OAC$ and $\Delta \, OBC$ are right angled, so $$OA = OC \cos\Big(\angle \, AOC \Big) = c \cos(\alpha) \,\,\, \text{ and } \,\,\, OB = OC \cos\Big(\angle \, BOC \Big) = c \cos(\beta)$$
The point $H$ is the orthogonal projection of $C$ onto the horizontal plane $OAB$. Thus, segment $CH$ is orthogonal to the plane $OAB$. By construction, $OA \, \perp \, CA$ and $OA \, \perp \, CH$, which means that $OA$ is perpendicular to the whole plane $ACH$ and in particular $OA \, \perp \, HA$. Analogously, $OB \, \perp \, HB$.
So we want to find $h = CH$. Since $CH$ is perpendicular to the plane $OAB$, and in particular $CH \, \perp \, OH$, the triangle $\Delta\, OHC$ is right angled so by Pythagoras' theorem $$h = CH = \sqrt{\, CH^2 - OH^2 \,} = \sqrt{\, c^2 - OH^2 \,}$$ so our goal would be to calculate $OH^2$. Since $$\angle \, OAH = 90^{\circ} = \angle \, OBH$$ the quadrilateral $OAHB$ is inscribed in a circle and $OH$ is the diameter of that circumcircle. By the law of sines $$OH = \frac{AB}{\sin\Big(\angle \, AOB\Big)} = \frac{AB}{\sin(\gamma)}$$ and thus $$OH^2 = \frac{AB^2}{\sin^2(\gamma)}$$ Next, apply the law of cosines to the triangle $\Delta \, OAB$ $$AB^2 \, = \, OA^2 \, + \, OB^2 \, - \, 2 \cdot OA \cdot OB\cdot \cos(\gamma)$$ $$AB^2 \, = \, c^2 \cos^2(\alpha) \, + \, c^2 \cos^2(\beta) \, - \, 2 c^2 \cos(\alpha) \cos(\beta) \cos(\gamma)$$ $$OH^2 = \frac{c^2 \big( \, \cos^2(\alpha) \, + \cos^2(\beta) - 2 \cos(\alpha) \cos(\beta) \cos(\gamma) \,)}{\sin^2(\gamma)}$$ $$h = CH = \sqrt{c^2 - OH^3} = \sqrt{\, c^2 \, - \, \frac{c^2 \big( \, \cos^2(\alpha) \, + \cos^2(\beta) - 2 \cos(\alpha) \cos(\beta) \cos(\gamma) \,\big)}{\sin^2(\gamma)} \,}$$ so after some term rearrangement $$h = c \, \frac{\sqrt{\,1 + 2\, \cos(\alpha) \cos(\beta) \cos(\gamma) - \cos^2(\alpha) - \cos^2(\beta) - \cos^2(\gamma)\,}}{\sin(\gamma)}$$
A vector solution. As before, you know the angles between the three pairs of vectors, but you know only the length of vector $\vec{c}$. Then, let $\hat{a} = \frac{1}{a} \, \vec{a}$ and let $\hat{b} = \frac{1}{b} \, \vec{b}$ be the two unit vectors aligned with vectors $\vec{a}$ and $\vec{b}$ respectively.
Our next step is to construct a unit vector $\hat{e}$ that is in the plane $OAB$ and is orthogonal to $\hat{a}$. Then, vector $\hat{b}$ should be a linear combination $$\hat{b} = \lambda \hat{a} + \mu \hat{e}$$ Then, \begin{align} &\hat{b}\cdot\hat{b} = \big( \lambda \hat{a} + \mu \hat{e} \big) \cdot \big( \lambda \hat{a} + \mu \hat{e} \big) = \lambda^2 (\hat{a}\cdot\hat{a}) + 2\lambda \mu (\hat{a}\cdot \hat{e}) + \mu^2(\hat{e}\cdot \hat{e}) \\ &\hat{a} \cdot \hat{b} = \lambda (\hat{a}\cdot \hat{a}) + \mu (\hat{a} \cdot \hat{e})\\ &\hat{e} \cdot \hat{b} = \lambda (\hat{e}\cdot \hat{a}) + \mu (\hat{e} \cdot \hat{e}) \end{align} so when you rewrite the dot products, having in mind that the vectors involved are unit and $\hat{a}$ and $\hat{e}$ are orthogonal, i.e. $\hat{a}\cdot \hat{e} = 0$ \begin{align} &1 = \lambda^2 + \mu^2\\ &\cos(\gamma) = \lambda\\ &\hat{e} \cdot \hat{b} = \mu \end{align} which means that $\mu = \sin(\gamma)$ and thus $$\hat{b} = \cos(\gamma) \hat{a} + \sin(\gamma) \hat{e}$$ and consequently $$\hat{e} = \frac{\hat{b} - \cos(\gamma) \hat{a}}{\sin(\gamma)} $$ Finally, consider the cross product
$$\hat{a} \times\hat{e} = \hat{a} \times \left(\frac{\hat{b} - \cos(\gamma) \hat{a}}{\sin(\gamma)}\right) = \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}$$
which is a third unit vector, orthogonal to $\hat{a},\, \,\hat{b}$ and $\hat{e}$, i.e. to the whole plane $OAB$.
The three unit pairwise orthogonal vectors $\hat{a}, \, \hat{e} \, \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}$ are linearly independent in 3D, so there unique three numbers $c_a, \, c_e, \, h$ for which the vector $\vec{c}$ can be expressed as $$\vec{c} = c_a \hat{a} + c_e \hat{e} + h \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}$$ Having in mind that since the three unit vectors $\hat{a}, \, \hat{e} \, \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}$ are pairwise orthogonal, their pairwise dot products are zero, which means that we can dot product $\vec{c}$ with each of the three vectors $\hat{a}, \, \hat{e} \, \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}$ and find the coefficients: \begin{align} &\vec{c} \cdot \hat{a} = \left(c_a \hat{a} + c_e \hat{e} + h \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}\right) \cdot \hat{a} = c_a (\hat{a}\cdot\hat{a}) + c_e (\hat{e}\cdot\hat{a}) + h \left(\frac{\hat{a} \times \hat{b}}{\sin(\gamma)} \cdot\hat{a}\right) = c_a\\ &\vec{c} \cdot \hat{e} = \left(c_a \hat{a} + c_e \hat{e} + h \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}\right) \cdot \hat{e} = c_a (\hat{a}\cdot\hat{e}) + c_e (\hat{e}\cdot\hat{e}) + h \left(\frac{\hat{a} \times \hat{b}}{\sin(\gamma)} \cdot\hat{e}\right) = c_e\\ &\vec{c} \cdot \left(\frac{\hat{a} \times \hat{b}}{\sin(\gamma)} \right) = \left(c_a \hat{a} + c_e \hat{e} + h \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}\right) \cdot \left(\frac{\hat{a} \times \hat{b}}{\sin(\gamma)} \right) = h\\ \end{align} Hence \begin{align} &c_a = \vec{c} \cdot \hat{a} = c \, \cos(\alpha)\\ &c_e = \vec{c} \cdot \hat{e} = \vec{c} \cdot \left(\frac{\hat{b} - \cos(\gamma) \hat{a}}{\sin(\gamma)}\right) = \frac{(\vec{c} \cdot \hat{b}) - \cos(\gamma) (\hat{c}\cdot \hat{a})}{\sin(\gamma)} = \frac{c\, \cos(\beta) - c\, \cos(\gamma) \cos(\alpha)}{\sin(\gamma)} \end{align} Now, by the 3D Pythagoras' theorem $$c^2 = c_a^2 + c_e^2 + h^2$$ we can express $h^2$ as $$h^2 = c^2 - c^2 \, \cos^2(\alpha) - c^2 \, \left(\frac{\cos(\beta) - \cos(\gamma) \cos(\alpha)}{\sin(\gamma)}\right)^2$$ and so $$h = c \, \sqrt{ \, 1 - \cos^2(\alpha) - \, \left(\frac{\cos(\beta) - \cos(\gamma) \cos(\alpha)}{\sin(\gamma)}\right)^2 \, }$$ and after some rearrangement of trigonometric terms you get the same result as before.