A Geometric Application Of Differentiation

Solving $\theta$ in terms of $x$ :

$\theta = \arctan{\frac{30}{x}}-\arctan{\frac{10}{x}}$

Differentiating with respect to $x$ and setting equal to zero to find the maximum for $\theta$ :

$0 = \frac{-30}{x^2+900}+\frac{10}{x^2+100}$ , $x = \sqrt{300} = \boxed{10\sqrt{3}}$

Plugging this value of $x$ into the function for $\theta$ to solve for the largest angle:

$\theta = \arctan{\sqrt{3}}-\arctan{\frac{1}{\sqrt{3}}} = \frac{\pi}{6} = \boxed{30˚}$

Consider the circle passing through $A$ and $B$ and tangent to the line $CE$ .

Then consider any other circle passing through $A$ , $B$ and cutting the line $CE$ at two distinct points $D$ and $E$ .

First we observe that any angle of viewing can be considered as the angle in the segment of the circumcircle of the triangle $ABX$ where $X$ is the point on $CE$ where the observer is standing.

So all we have to do is optimize the angle in the segment subtended by all such circles.

Now we check that

$m(AEB)=m(ADB)=m(AFB)$ ( as they are angles in the same segment of the bolder circle)

and $m(AFB)=m(AGB)-m(FAG)<m(AGB)$

so that we have $m(AEB)<m(AGB)$

(All the circles we consider should cut the circle at one point at least because the point(s) of intersection of the circle and the line $CE$ denote the place where the person is standing)

So for all circles, the maximum angle is achieved when it is tangent to the line $CE$ .

Now we use the power of point $C$ with respect to the circle passing through $ABG$ to get,

$CA.CB=CG^2$ , which yields $x=10\sqrt{3}$ .

Now by alternate-segment theorem, we notice that $m(CGB)=m(BAG)$ .

Also, in the right angled triangle $GBC$ , we know the side lengths $CB=10$ and $CG=10\sqrt{3}$ so that we know the value of $m(CGB)=\frac{\pi}{6}$ and we have by Pythagoras' Theorem, $GB=20=AB$ , so that $BAG$ is an isosceles triangle with $m(AGB)=m(BAG)=m(BGC)=\frac{\pi}{6}$ .

Hence proved.

P.S. This problem is a beautiful example of how geometry can assist calculus.