How would I calculate the area of a rectangle on a sphere using vertical and horizontal angles? Imagine a sphere being one's eyeball and the rectangular area being the picture of one's view.  Like putting a name tag sticker on a balloon. How can I find the area of the rectangle on the sphere?
 A: You need to define a spherical rectangle. There is no figure on a sphere with four sides, each a segment of a great circle (a straight line on the sphere) and with four right angles. 
A plane rectangle has opposite sides equal in length and all angles  equal. So, let's define a spherical rectangle in the same way:
Definition: a rectangle on the sphere is a four-sided polygon with opposite sides having equal length and all four internal angles of equal magnitude.
For example, take the two meridians on the globe at longitude +15 deg and -15 deg. Join the points at latitude +20 deg on each meridian by a great circle segment (straight line). Do the same for the points at latitude -20 deg on each meridian. Voila: we have a spherical rectangle.
To get the area, draw either diagonal, calculate the area of either resulting triangle and double the value.
A: 
Let’s describe your “rectangle” this way:
Think of your figure as centered on the point $0$ longitude, $0$ latitude, extending upwards and downwards by $\alpha$ and to the right and to the left by $\beta$. These are angles, as you realize. Now the top side is perpendicular to the vertical line segment, similarly the bottom side. In the same way, the left and right sides are perpendicular to the horizontal line. We need look only at the upper left quadrangle, which has three right angles and an unknown angle $C$, whose size we need to evaluate, as @Rocky pointed out in his response.
It’s now a question of spherical trigonometry, which seems to have been understood already by Napier, who died a quarter-century before Newton’s birth. The first task is to find the length of the segment I’ve labeled $\gamma$, and that’s nicely given by the Spherical Pythagorean Theorem, it says $\cos\gamma=\cos\alpha\cos\beta$. We also need the angles $A$ and $B$, and they’re given by another of Napier’s rules for spherical right triangles, namely that $\tan(\text{angle})=\tan(\text{opposite})/\sin(\text{adjacent})$, so that here, $\tan A=\tan\alpha/\sin\beta$, and similarly, $\tan B=\tan\beta/\sin\alpha$.
Of course it’s the complements of these angles that we want for solving the triangle containing the angle $C$, I’ll call them $\bar A$ and $\bar B$. But $\tan A=\cot \bar A=1/\tan \bar A$. Thus we have the relations $\tan\bar A=\sin\beta\cot\alpha$, $\tan\bar B=\sin\alpha\cot\beta$. You can express $\bar A$ and $\bar B$ as the values of the inverse tangent function, if you like.
Now finally, we have a triangle with data of ASA type, and for this you use the Dual Law of Cosines, which says, for a triangle labeled in the standard way with angles $A$, $B$, and $C$ with opposite sides $a$, $b$, and $c$ respectively, that
$$
\cos C=-\cos A\cos B +\sin A\sin B\cos c\,.
$$
For our data, replace $A$ with $\bar A$, $B$ with $\bar B$, and $c$ with $\gamma$.
We do have the previous fact that $\cos\gamma=\cos\alpha\cos\beta$.
Let’s do a representative example. Suppose our rectangle extends $20^\circ$ top to bottom and $40^\circ$ left to right, so that $\alpha=10^\circ$ and $\beta=20^\circ$. Then $\gamma=\arccos(\cos10^\circ\cos20^\circ)=22.2687^\circ$. Continuing, $\bar A=\arctan(\sin20^\circ/\tan10^\circ)=62.7268^\circ$ and $\bar B=\arctan\sin10^\circ/\tan20^\circ=25.5056^\circ$. Finally,
\begin{align}
C&=\arccos(\sin62.7268^\circ\sin25.5056^\circ\cos22.2687^\circ
-\cos62.7268^\circ\cos25.5056^\circ)\\
&=93.4049^\circ
\end{align}
Now we muliply by $4$ and subtract $360^\circ$, then convert to radians, getting $.2377$, and to get an area quantity, multiply by $r^2$, the radius of your sphere, in other words the distance to the center of your picture.
