Determine the third angle of the triangular apparatus.

The sum of angles in a triangle is $180^\circ$. Given angles are $59^\circ$ and $73^\circ$.

The angle of the remaining vertex is $\boxed{\mathbf{48^\circ}}$.

Let the given side be $c = 12$ ft, opposite the $48^\circ$ angle. Let the other angles be $A = 59^\circ$ and $B = 73^\circ$, with opposite sides $a$ and $b$ respectively.
The Sine Rule states: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
We have $\frac{a}{\sin 59^\circ} = \frac{b}{\sin 73^\circ} = \frac{12}{\sin 48^\circ}$.

The length of one remaining side is $\boxed{\mathbf{13.84 \text{ ft}}}$.

The length of the other remaining side is $\boxed{\mathbf{15.44 \text{ ft}}}$.

The equation is $5 \sec \theta - 2 \sec^2 \theta = \tan^2 \theta - 1$.
Recall the identity $\sec^2 \theta = 1 + \tan^2 \theta$, which implies $\tan^2 \theta = \sec^2 \theta - 1$.
Substitute $\tan^2 \theta - 1$ with $(\sec^2 \theta - 1) - 1$:

Move all terms to one side to form a quadratic equation in terms of $\sec \theta$:

Let $x = \sec \theta$. The equation becomes $3x^2 - 5x - 2 = 0$.
Factor the quadratic equation:

This gives two possible values for $x$:

So, $\sec \theta = -\frac{1}{3}$ or $\sec \theta = 2$.

Since $\sec \theta = \frac{1}{\cos \theta}$, we have $\cos \theta = \frac{1}{\sec \theta}$.
Case 1: $\cos \theta = \frac{1}{-1/3} = -3$. This is not possible because the range of $\cos \theta$ is $[-1, 1]$. So, there are no solutions from this case.
Case 2: $\cos \theta = \frac{1}{2}$.
The principal value (reference angle) for $\cos \theta = \frac{1}{2}$ is $\alpha = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ$.
Since $\cos \theta$ is positive, $\theta$ lies in Quadrant I and Quadrant IV.
In Quadrant I: $\theta_1 = \alpha = 60^\circ$.
In Quadrant IV: $\theta_2 = 360^\circ - \alpha = 360^\circ - 60^\circ = 300^\circ$.
The solution set for $\theta$ is $\boxed{\mathbf{\{60^\circ, 300^\circ\}}}$.

The formula for $\tan(X - Y)$ is $\tan(X - Y) = \frac{\tan X - \tan Y}{1 + \tan X \tan Y}$.
Substitute $X = 45^\circ$ and $Y = 2A$:

We know that $\tan 45^\circ = 1$.

The formula for $\tan 2A$ is $\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}$.

This matches the expression given by the student, thus showing how it holds.