# 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?

• If you prefix the names of trigonometric functions with a backslash in LaTeX/MathJax, they will be rendered in upright font with appropriate spacing instead of an chain of italic letters. Please compare cot\alpha → $cot\alpha$ to \cot\alpha → $\cot\alpha$. May 4, 2022 at 13:24

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,

$$h^2 = \dfrac{ p^2 + q^2 + r^2 \pm \sqrt{3} \sqrt{2 p^2 q^2 + 2 p^2 r^2 + 2 q^2 r^2 - p^4 - q^4 - r^4} }{ 2 (p^4 + q^4 + r^4 - p^2 q^2 - p^2 r^2 - q^2 r^2)}$$