#1**+1 **

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"

Melody Oct 24, 2019

#2**+3 **

Melody, I interpret this as follows:

Imagine A,B and C in a straight line on the ground, with B in the middle. Now picture a vertical plane (not aeroplane!) intersecting that line. That plane does not go through the aeroplane, i.e. the aeroplane is off to one side of the vertical plane.

Does that make it clearer?

Alan
Oct 24, 2019

#4**+4 **

**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:**

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

heureka Oct 24, 2019

#7**+1 **

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

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

OldTimer Oct 25, 2019

#8**+1 **

Not quite,

Let the point where the plane is be P

F is the point on the ground directly below.

Obviously you must assume that the ground is flat so A,B,C, and F all are in the same plane

This also means that angles PFA, PFB, PFC are all 90 degrees.

This is becasue PF is vertical and AF, BF, and CF are all lines on the horizontal ground.

also,

r, s and t are ground distances to F (which is directly below the plane.)

__Reason for edit.__

I have to edit becasue of display errors, I think this happens when I use the less than sign to donote angles ...

Melody
Oct 25, 2019