A tower stands vertically in the interior of a field which has the shape of an equilateral triangle of side $a$. If the angles of elevation... 
A tower stands vertically in the interior of a field which has the shape of an equilateral triangle of side $a$. If the angles of elevation of the top of the tower are $\alpha$,$\beta$,$\gamma$ from the corners of the field find the height of the tower.

Answer: $$a\sqrt{\frac{p^2+q^2+r^2-\sqrt{2p^2q^2+2p^2r^2+2q^2r^2 -p^4 -q^4 - r^4}}{2(p^4 + q^4 + r^4 -p^2q^2 -p^2r^2 - q^2r^2)}}$$ where $p= \cot \alpha$, $q = \cot \beta$ and $r = \cot \gamma$

My Attempt
$[.]$ denotes area
$$[APB]+[BPC]+[CPA] = [ABC]$$
$$[ABC] = \frac{\sqrt{3}}{4}a^2$$
I tried finding the area of $\triangle APB$,$\triangle BPC$,$\triangle CPA$ by Heron's formula but the result was too complicated because I had to add square roots : something of the form $\sqrt{x} + \sqrt{y} + \sqrt{z} = w$. Moreover, the $x,y,z$ in this expression are also complicated.
I admit that the answer is complicated too but I am searching for an elegant approach towards the solution (something like substitution or anything else). Is there any elegant approach?

As the figure shows the distances between the base of the tower $P$ and each of the three vertices $A,B,C$ of equilateral $\triangle ABC$ is
$ PA = h \cot \alpha = h p $
$ PB = h \cot \beta = h q $
$ PC = h \cot \gamma = h r $
Now using Barycentric coordinates, and taking $A$ to be the origin, we can express point $P$ in terms of $B $ and $C$ as follows
$ P = c_1 B + c_2 C $
so that
$ AP = P = c_1 B + c_2 C , BP = (c_1 - 1) B + c_2 C , CP = c_1 B + (c_2 - 1) C $
I'll assume that the side length $a = 1$.  Taking the magnitude of these vectors, and noting that $|B| = |C| = 1$ and that $B \cdot C = \dfrac{1}{2} $, then we can write the following three equations
$c_1^2 + c_2^2 + c_1 c_2 = h^2 p^2 \hspace{10pt} (*) $
$(c_1 - 1)^2 + c_2^2 + (c_1 - 1) c_2 = h^2 q^2 $
$ c_1^2 + (c_2 - 1)^2 + c_1 (c_2 - 1) = h^2 r^2 $
We want to eliminate $c_1, c_2$, so subtract each pair of equations, i.e. (1)-(2) , and (1) - (3), we get
$ 2 c_1 + c_2 = h^2 (p^2 - q^2) + 1 $
$ c_1 + 2 c_2 = h^2 (p^2 - r^2) +1$
Solving this $2 \times 2$ system, gives us
$ c_1 = \dfrac{1}{3} (  h^2 (p^2 - 2 q^2 + r^2 ) + 1 )$
$ c_2 = \dfrac{1}{3} ( h^2 (p^2 - 2 r^2 + q^2 ) + 1 )$
substituting $c_1, c_2$ into Eq. $(*)$ and expanding, gives us
$  h^4 K + h^2 L + M = 0  $
where
$K = p^4 + q^4 + r^4 - p^2 q^2 - p^2 r^2 - q^2 r^2 $
$ L = -  ( p^2 + q^2 + r^2 ) $
$ M = 1 $
Combining the $h^2 $ coefficients, we get the quadratic equation (in $h^2$)
From this, using the quadratic formula, the solution is
$ h^2 = \dfrac{ -L \pm \sqrt{ L^2 - 4 K } }{ 2 K } $
we have
$ - L = p^2 + q^2 + r^2 $
and
$L^2 - 4 K = p^4 + q^4 + r^4 + 2 p^2 q^2 + 2 p^2 r^2 + 2 q^2 r^2 - 4 (p^4 + q^4 + r^4 - p^2 q^2 - p^2 r^2 - q^2 r^2) $
And this reduces to
$ L^2 - 4 K = 3 ( 2 p^2 q^2 + 2 p^2 r^2 + 2 q^2 r^2 - p^4 - q^4 - r^4 ) $
Hence,
$ h^2 = \dfrac{ p^2 + q^2 + r^2 \pm \sqrt{3} \sqrt{2 p^2 q^2 + 2 p^2 r^2 + 2 q^2 r^2 - p^4 - q^4 - r^4} }{ 2 (p^4 + q^4 + r^4 - p^2 q^2 - p^2 r^2 - q^2 r^2)}$
