Distance from $A$ to $\triangle BCD$ if $\angle BAC=\angle CAD=\angle DAB=60^\circ$? Let $A$, $B$, $C$, and $D$ be four points in space such that
$$\angle BAC=\angle CAD=\angle DAB=60^\circ.$$
If $AB=1$, $AC=2$, and $AD=6$, then what is the distance between $A$ and the plane of $\triangle BCD$?

Observations: $ABC$ and $BCD$ are right, as $ABC$ is $30-60-90$ and then using Law of Cosines we han find the other side lengths for $BCD.$ I don't know how to use this to solve for the distance desired though. I have tried letting this distance be $h$ and setting up equations.
 A: You can use the formula for tetrahedron volume (wiki),
$V = \frac{abc}{6} \sqrt{1+2 \cos \alpha \cos \beta \cos \gamma - \cos^2 \alpha - \cos^2 \beta - \cos^2 \gamma}$
$\alpha, \beta, \gamma$ are angles between edges at a vertex and $a, b, c$ are lengths of edges from the vertex.
Here we know that at vertex $A$, $\angle BAC = \angle BAD = \angle CAD = 60^0$ and $AB = 1, AC = 2, AD = 6$.
So $V = \displaystyle \frac{1\cdot2\cdot6}{6} \sqrt{1 + \frac{2}{2^3} - \frac{3}{2^2}} = \sqrt2$
We also know that the volume of Tetrahedron is given by $V = \frac{1}{3} A \cdot h$ where $A$ is the area of the base and $h$ is the altitude.
As you found out, $\triangle BCD$ is right angled triangle with $BC = \sqrt3, CD = \sqrt{28}, BD = \sqrt{31}$. So, $A = \sqrt{21}$.
So altitude from $A$ to base BCD, $h = \frac{3V}{A} = \sqrt{\frac{6}{7}}$
A: 
Let $\theta$ be the angle between the planes $ABC$ and $BCD$.
Recognize the right triangles $ABC$ and $BCD$ to establish the distance equation
$$AD^2 = (AB+ CD \cos\theta)^2 + BC^2 +(CD\sin\theta)^2
$$
which leads to
$\cos\theta =\frac1{\sqrt7}$.
Then, the distance from $A$ to the plane $BCD$ is
$$d= AB \sin\theta = \sqrt{\frac67}$$
A: You can find the volume of the tetrahedron from the lengths of the edges from this Heron-like formula:
$$
{\displaystyle V={\frac {\sqrt {4a^{2}b^{2}c^{2}-a^{2}X^{2}-b^{2}Y^{2}-c^{2}Z^{2}+XYZ}}{12}}}
$$
where
$a=AB$, $b=AC$, $c=AD$,
$x=CD$, $y=BD$, $z=BC$
and
$$
{\displaystyle {\begin{aligned}X&=b^{2}+c^{2}-x^{2},\\Y&=a^{2}+c^{2}-y^{2},\\Z&=a^{2}+b^{2}-z^{2}.\end{aligned}}}
$$
Just divide the triple of that volume by the area of $BCD$ to find the relative height $AH$.
A: I see from comment that you want to avoid trigonometry: so you can follow these steps.

*

*The edges from A form the same angle between them. We can take the z axis from A to be symmetrically  placed wrt the three edges. Then they project on the xy plane at $120^\circ$.


*Take the three unitary vectors on the xy plane and add the same z component $h$.
Take the dot product and determine $h$ for the angle to be $60^\circ$.


*Determined $h$ you have the three unit vectors along the edges. Multiply by $1,2,6$ and get the vertices $B,C,D$.


*Find the plane through $B,C,D$ and the distance of $A$ from it, or work it out through the determinant of the base (area), and the determinant of the tetrahedron (volume).
A: Start with a regular tetrahedron AEFD of edge length 6. Choose B on AE and C on AF at the correct distance. As the ratio AC:AB is 2, BC is perpendicular to AE. D dropped perpendicular to AEF cuts the height over AE of triangle AEF 2:1, the same ratio as C. In other words D lies in the plane through C perpendicular to the height which is parallel to BC, hence DC is perpendicular to BC. The volume of ABCD is 1/18 that of AEFD since that is the ratio of areas ABC and AEF. It is also 1/6 of the product BC x CD x d where d is the distance we are looking for. The volume of regular tetrahedron AEFD is $18 \times \sqrt{2}$, The lengths BC and CD are (here we need a little bit of Pythagoras) $\sqrt{3}$ and $2 \times \sqrt{7}$. Therefore d = $\sqrt{6/7}$.
