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# Heights and Distances

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10 Can you help...thanks & regards

Oct 24, 2019
edited by Guest  Oct 24, 2019
edited by OldTimer  Oct 24, 2019
edited by OldTimer  Oct 24, 2019

#1
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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"

Oct 24, 2019
edited by Melody  Oct 24, 2019
#2
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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
edited by Alan  Oct 24, 2019
#3
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Thanks Alan, Yes that helps me understand the question.

It probably doesn't help me answer it but it gives me a starting point. Melody  Oct 24, 2019
#4
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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:

$$\begin{array}{|lrcll|} \hline (1) & t^2 &=& s^2+c^2-2sc\cos(\epsilon) \\\\ & r^2 &=& s^2+c^2-2sc\cos(180^\circ-\epsilon) \\ &&& \boxed{ \cos(180^\circ-\epsilon)=- \cos(\epsilon)} \\ (2) & r^2 &=& s^2+c^2+2sc\cos(\epsilon) \\ \hline (2)+(1): & r^2+t^2 &=& s^2+c^2+2sc\cos(\epsilon)+s^2+c^2-2sc\cos(\epsilon)\\ & r^2+t^2 &=& 2s^2+2c^2 \\ & r^2+t^2-2s^2 &=& 2c^2\\ &&& \boxed{ r^2=h^2\cot^2(\alpha)\\ s^2=h^2\cot^2(\beta)\\ t^2=h^2\cot^2(\gamma) }\\ & h^2\cot^2(\alpha)+h^2\cot^2(\gamma)-2h^2\cot^2(\beta) &=& 2c^2 \\ & h^2\left(\cot^2(\alpha)+ \cot^2(\gamma)-2 \cot^2(\beta)\right) &=& 2c^2 \\ & h^2 &=& \dfrac{2c^2 }{ \cot^2(\alpha)+ \cot^2(\gamma)-2 \cot^2(\beta) } \\\\ &\mathbf{ h } &=& \mathbf{ \dfrac{c\sqrt{2} }{ \sqrt{\cot^2(\alpha)+ \cot^2(\gamma)-2 \cot^2(\beta)} } } \\ \hline \end{array}$$ Oct 24, 2019
#5
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Nice work Heureka!

Melody  Oct 24, 2019
#6
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Thank you, Melody ! heureka  Oct 24, 2019
#7
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Just one point...can you confirm that I got my geometry right as follows: Oct 25, 2019
edited by OldTimer  Oct 25, 2019
#8
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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
edited by Melody  Oct 25, 2019
#9
+1 Thanks for the clarification!

Oct 25, 2019
#10
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Yes that looks right.

And h is the height above ground that the plane is flying

FP =  h units

Melody  Oct 25, 2019
edited by Melody  Oct 25, 2019