Hello, ihmth!

I don't understand the diagram for the first one . . .

2) The lower edge of a picture is $\displaystyle a$ ft, the upper edge is $\displaystyle b$ ft above the eye of an observer.

At what horizontal distance shall he stand

if the vertical angle subtended by the picture is the greatest? I have a rather unplesant solution . . .

Code:

* P
* | :
* | :
* | :
* | b-a
* | :
* | :
* | :
* * Q
* * | :
* θ * | a
* * | :
E * - - - - - - - - - - - * R
x

$\displaystyle E$ is the eye of the observer.

$\displaystyle PQ$ is the height of the picture.

. . $\displaystyle a = QR,\;\;b = PR,\;\;PQ \:=\: b-a$

Let $\displaystyle x = ER$, the horizontal distance.

Let $\displaystyle \theta = \angle PEQ$

In right triangle $\displaystyle QRE\!:\;\;EQ \:=\:\sqrt{x^2+a^2}$

In right triangle $\displaystyle PRE\!:\;\;EP \:=\:\sqrt{x^2+b^2}$

In $\displaystyle \Delta PEQ$, use the Law of Cosines: .$\displaystyle \cos\theta \:=\:\frac{EQ^2 + EP^2 - PQ^2}{2(EQ)(EP)}$

We have: .$\displaystyle \cos\theta \;=\;\frac{(x^2+a^2) + (x^2+b^2) - (b-a)^2}{2\sqrt{x^2+a^2}\sqrt{x^2+b^2}}$

. . which simplifies to: .$\displaystyle \cos\theta \;=\;\frac{x^2+ab}{\sqrt{(x^2+a^2)(x^2+b^2)}} $

And I let __you__ maximize $\displaystyle \theta$ . . .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I have an alternate equation: .$\displaystyle \tan\theta \;=\;\frac{(b-a)x}{x^2+ab}$

. . but its derivation is even harder to explain.