Hello, jjdemick!

Here's the first part . . .

1. A sign on top of a 50-foot tall building is 10 feet tall.

The company president asked you to find the best view of the sign,

with pedestrians standing currently about 50 feet and are trampling

the company's lawn because the sidewalk is 60 feet away from the building.

The angle between the lines of sight from the observer's eye to the top

and bottom of the sign is the measure of the observer's view of the sign,

and best view occurs when the angle is the biggest.

The observer's eyes are 5 feet from the ground (observer is 5' 3" tall).

(a) Determine that, when the sign is 10 ft, the ideal distance is 49.7 ft. Code:

A *
| *
10| *
| *
B * *
| * θ *
45| * *
| α * *
C * - - - - - - - * P
5| x |5
D * - - - - - - - * Q
x

The flagpole is: $\displaystyle AB = 10$

The building is: $\displaystyle BD = 50$

The observer is: $\displaystyle PQ = 5$.

. . Hence: $\displaystyle CD = 5,\:BC = 45$

The observer's distance from the building is: $\displaystyle DQ = x = CP$

Let $\displaystyle \theta = \angle APB,\;\alpha = \angle BPC$

In right triangle $\displaystyle BCP\!:\;\tan\alpha \,=\,\frac{45}{x}$ .[1]

In right triangle $\displaystyle ACP\!:\;\tan(\theta + \alpha) \:=\:\frac{55}{x} \quad\Rightarrow\quad \frac{\tan\theta + \tan\alpha}{1 - \tan\alpha\tan\theta} \:=\:\frac{55}{x}$

Substitute [1]: .$\displaystyle \frac{\tan\theta + \frac{45}{x}}{1 - \frac{45}{x}\tan\theta} \:=\:\frac{55}{x}$

Multiply by $\displaystyle \frac{x}{x}\!:\;\;\frac{x\tan\theta + 45}{x - 45\tan\theta} \:=\:\frac{55}{x} \quad\Rightarrow\quad x^2\tan\theta + 45x \:=\:55x - 2475\tan\theta $

Then: .$\displaystyle x^2\tan\theta + 2475\tan\theta \:=\:10x \quad\Rightarrow\quad (x^2+2475)\tan\theta \:=\:10x$

. . Hence: .$\displaystyle \tan\theta \;=\;\frac{10x}{x^2+2475}$ . . . and we will maximize $\displaystyle \theta.$

Differentiate: .$\displaystyle \sec^2\!\theta\,\frac{d\theta}{dx} \;=\;10\cdot\frac{2475-x^2}{(x^2+2475)^2} \quad\Rightarrow\quad \frac{d\theta}{dx} \;=\;10\cos^2\!\theta\cdot\frac{2475-x^2}{(x^2+2475)^2} \;=\;0 $

. . If $\displaystyle \cos\theta = 0$, then: .$\displaystyle \theta \,=\,\pm90^o$ . . . impossible

. . Therefore: .$\displaystyle 2475-x^2\:=\:0 \quad\Rightarrow\quad x \:=\:\sqrt{2475} \:\approx\:\boxed{49.7\text{ ft}} $