MHB When is the Maximum Angle for Viewing a Tower from an Airplane?

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Hey! :giggle:

An aeroplane flies over a tower of height $h> 0$ at height $H> h$. At what distance $x$ is the angle $\alpha$ at which the tower is seen from the aeroplane, maximum?
(You can use elementary geometry and that $\arctan'(x)=\frac{1}{1+x^2}$.)

1612819439378.png
From Pythagorean Theorem for the larger triangle we have that $H^2+x^2=:y^2$.

Do we apply for the smaller triangle law of cosine? But the upper side of that triangle is not known,and not related to the other triangle.

From the hint... Do we have to use somehow $\tan(\alpha)$ ?

:unsure:
 
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Hey mathmari!

Suppose we use only tangents.
Which rectangular triangles can we find to apply the tangent to? 🤔
 
Klaas van Aarsen said:
Suppose we use only tangents.
Which rectangular triangles can we find to apply the tangent to? 🤔

We have the below rectangular triangle $ABC$ :

1612820733087.png


So we get $\tan (\phi)=\frac{x}{AB}$, right?

But we have to relate it with angle $\alpha$, right? Or do we have to consider an other triangle?

:unsure:
 
mathmari said:
So we get $\tan (\phi)=\frac{x}{AB}$, right?

But we have to relate it with angle $\alpha$, right? Or do we have to consider an other triangle?
AB is the same as H isn't it? :unsure:

And yes, can we find another rectangular triangle? 🤔
 
Klaas van Aarsen said:
AB is the same as H isn't it? :unsure:

And yes, can we find another rectangular triangle? 🤔

Do we take a line that passes through $C$ and the intersection point with the upper horizinal line is say $D$ (then $|CD|=H$) and we get the triangle $ACD$ ?

:unsure:
 
mathmari said:
Do we take a line that passes through $C$ and the intersection point with the upper horizinal line is say $D$ (then $|CD|=H$) and we get the triangle $ACD$ ?
That triangle $ACD$ is congruent with $ABC$ isn't it? :unsure:

Can we find another rectangular triangle?
Perhaps if we draw extra help lines? 🤔
 
Klaas van Aarsen said:
Can we find another rectangular triangle?
Perhaps if we draw extra help lines? 🤔

I suppose that we have to use the two line that have the angle $\alpha$ in between, or not? :unsure:
 
mathmari said:
I suppose that we have to use the two line that have the angle $\alpha$ in between, or not?
How about these triangles:
\begin{tikzpicture}[scale=2]
\coordinate[label=left:A] (A) at (0,3);
\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=below:C] (C) at (4,0);
\coordinate[label=T] (T) at (4,1);
\coordinate[label=above right:T'] (Ta) at (0,1);
\draw[thick] (A) -- (B) -- (C) -- cycle;
\draw[thick] (A) -- (T) -- node[ right ] {h} (C) (Ta) -- (T);
\path (A) -- node[ left ] {H} (B) -- node[below] {x} (C) -- (A) (B) -- node[ right ] {h} (Ta);
\end{tikzpicture}
🤔
 
Klaas van Aarsen said:
How about these triangles:
\begin{tikzpicture}[scale=2]
\coordinate[label=left:A] (A) at (0,3);
\coordinate[label=left:B] (B) at (0,0);
\coordinate[label=below:C] (C) at (4,0);
\coordinate[label=T] (T) at (4,1);
\coordinate[label=above right:T'] (Ta) at (0,1);
\draw[thick] (A) -- (B) -- (C) -- cycle;
\draw[thick] (A) -- (T) -- node[ right ] {h} (C) (Ta) -- (T);
\path (A) -- node[ left ] {H} (B) -- node[below] {x} (C) -- (A) (B) -- node[ right ] {h} (Ta);
\end{tikzpicture}
🤔

Considering the triangle $AT'T$ we get $\tan (\angle T'AT)=\frac{x}{H-h}$.
Considering the triangle $ABC$ we get $\tan (\angle BAC)=\frac{x}{H}$.

We have that $\alpha=\angle T'AT-\angle BAC$. So we get $$\tan (\alpha)=\tan \left (\angle T'AT-\angle BAC \right )=\frac{\tan (\angle T'AT)-\tan (\angle BAC)}{1+\tan (\angle T'AT)\tan (\angle BAC)}=\frac{\frac{x}{H-h}-\frac{x}{H}}{1+\frac{x}{H-h}\frac{x}{H}}=\frac{\frac{xH-xH+xh}{H(H-h)}}{\frac{H(H-h)+x^2}{H(H-h)}}=\frac{xh}{H(H-h)+x^2}$$

Then we have that $\alpha(x)=\arctan \left (\frac{xh}{H(H-h)+x^2}\right )$.

To get the maximum $\alpha$ we calculate the roots of the first derivative :
$$\alpha'(x)=0 \Rightarrow \arctan' \left (\frac{xh}{H(H-h)+x^2}\right )=0\Rightarrow \frac{1}{1+\left (\frac{xh}{H(H-h)+x^2}\right )^2}=0$$

From thatequation we get the desiredvaluefor $x$. Is that correct? :unsure:
 
  • #10
Shouldn't we apply the chain rule? 🤔

Alternatively we might also observe that $\alpha(x)$ takes on its maximum value iff $\tan\alpha(x)$ does.
Then we don't need the derivative of $\arctan$ at all. 🤔

Btw, we can also write $\alpha = \arctan \angle T'AT - \arctan \angle BAC$. 🤔
 
  • #11
Klaas van Aarsen said:
Shouldn't we apply the chain rule? 🤔

Ahh yes.. So we have $$\alpha'(x)=0 \Rightarrow \arctan' \left (\frac{xh}{H(H-h)+x^2}\right )=0\Rightarrow \frac{1}{1+\left (\frac{xh}{H(H-h)+x^2}\right )^2}\cdot \left (\frac{xh}{H(H-h)+x^2}\right )'=0 $$ right? :unsure:
Klaas van Aarsen said:
Btw, we can also write $\alpha = \arctan \angle T'AT - \arctan \angle BAC$. 🤔

How do we get that? :unsure:
 
  • #12
mathmari said:
Ahh yes.. So we have $$\alpha'(x)=0 \Rightarrow \arctan' \left (\frac{xh}{H(H-h)+x^2}\right )=0\Rightarrow \frac{1}{1+\left (\frac{xh}{H(H-h)+x^2}\right )^2}\cdot \left (\frac{xh}{H(H-h)+x^2}\right )'=0 $$ right?
Yep. (Nod)

mathmari said:
How do we get that?
Oops. I meant $\alpha = \angle T'AT - \angle BAC = \arctan \frac x{H-h} - \arctan\frac x H$. :oops:
 
  • #13
Klaas van Aarsen said:
Yep. (Nod)Oops. I meant $\alpha = \angle T'AT - \angle BAC = \arctan \frac x{H-h} - \arctan\frac x H$. :oops:

Thank you! 🤩
 
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