Heights and Distances

Hi OldTimer :)

An aeroplane is observed at the same instant from three stations A,B, C in a horizontal straight line but not in a vertical plane through the aeroplane.

If AB=BC =c and the angles of elevation from A,B C are respectively  alpha, beta and gamma prove that the height of the plane is

\(\frac{c\sqrt2 }{\sqrt{(cot^2\alpha +cot^2\gamma -2cot^2\beta )} }\)

I have not got my head around what this question is asking.  Can you (or someone else) help with a pic?

this is the bit I am not understanding 

"a horizontal straight line but not in a vertical plane"

An aeroplane is observed at the same instant from three stations A,B, C in a horizontal straight line

but not in a vertical plane through the aeroplane.
If \(AB=BC =c\) and the angles of elevation from A,B C are respectively  \(\alpha\), \(\beta\) and \(\gamma\),

prove that the height of the aeroplane is

\(\dfrac{c\sqrt{2}}{\sqrt{\cot^2(\alpha)+\cot^2(\gamma)-2\cot^2(\beta) }}\)

\(\text{The height of the aeroplane $=h $} \\ \text{$F$ is the nadir point of the aeroplane } \\ \text{$r = \overline{AF}$ } \\ \text{$s = \overline{BF}$ } \\ \text{$t = \overline{CF}$ } \\ \text{$\angle CBF = \epsilon$ } \\ \text{$\angle ABF = 180^\circ-\epsilon$ } \)

\(\begin{array}{|rclcrcl|} \hline \cot(\alpha) &=& \dfrac{r}{h} &\text{or}& r &=& h\cot(\alpha) \qquad (1) \\ \cot(\beta) &=& \dfrac{s}{h} &\text{or}& s &=& h\cot(\beta) \qquad (2) \\ \cot(\gamma) &=& \dfrac{t}{h} &\text{or}& t &=& h\cot(\gamma) \qquad (3) \\ \hline \end{array} \)

cos-rule:

\(\begin{array}{|lrcll|} \hline (1) & t^2 &=& s^2+c^2-2sc\cos(\epsilon) \\\\ & r^2 &=& s^2+c^2-2sc\cos(180^\circ-\epsilon) \\ &&& \boxed{ \cos(180^\circ-\epsilon)=- \cos(\epsilon)} \\ (2) & r^2 &=& s^2+c^2+2sc\cos(\epsilon) \\ \hline (2)+(1): & r^2+t^2 &=& s^2+c^2+2sc\cos(\epsilon)+s^2+c^2-2sc\cos(\epsilon)\\ & r^2+t^2 &=& 2s^2+2c^2 \\ & r^2+t^2-2s^2 &=& 2c^2\\ &&& \boxed{ r^2=h^2\cot^2(\alpha)\\ s^2=h^2\cot^2(\beta)\\ t^2=h^2\cot^2(\gamma) }\\ & h^2\cot^2(\alpha)+h^2\cot^2(\gamma)-2h^2\cot^2(\beta) &=& 2c^2 \\ & h^2\left(\cot^2(\alpha)+ \cot^2(\gamma)-2 \cot^2(\beta)\right) &=& 2c^2 \\ & h^2 &=& \dfrac{2c^2 }{ \cot^2(\alpha)+ \cot^2(\gamma)-2 \cot^2(\beta) } \\\\ &\mathbf{ h } &=& \mathbf{ \dfrac{c\sqrt{2} }{ \sqrt{\cot^2(\alpha)+ \cot^2(\gamma)-2 \cot^2(\beta)} } } \\ \hline \end{array}\)

That's beautiful.....let's say....made my week!

Just one point...can you confirm that I got my geometry right as follows:

Thanks for the clarification!