Q:

Applying the law of sines

A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, $5$ mi. apart, to be $32°$ and $48°$.
(a) Find the distance of the plane from point $A$.
(the point with an angle of depression of $32°$)
(b) Find the elevation of the plane.
So drawing this out, I figured that the flight path of the plane and the highway would be parallel, thus angle A would be the alternate interior angle of the $32°$ angle of depression. The same would apply to angle B and the $48°$ angle of depression on its alternate side. 
So the third angle must be $100°$.
So for (a) I did the following:
$$\frac { 5 }{ sin(100) } =\frac { d }{ sin(48) } $$
$$\frac { (5)sin(48 }{ sin(100) } =\quad d\quad $$
$$d\quad \approx \quad 3.77\ mi. $$ 
and for (b) I did the following:
$$sin(32)=\frac { h }{ 3.77 } \quad $$
$$(3.77)sin(32)=h\quad $$
$$h\approx 2\quad mi. $$
I feel like my answers are based on the assumption that the angles truly are alternate interior ones. If that is not the case, then I am wrong and have no other way that I can think of to solve this. I'd like a hint in the right direction. Not the actual answers. 
Here is the picture of the problem:

A:

EDIT
Now by looking at the picture shown in the book, everything is clear:

First:

two lines (A) and (B) are parallel, and a third line (C) intersect them, the internal angles are equal, so your assumption is right.

Second:

in a triangle, the relation : $\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$ holds:

Third: the sum of angles of a triangle is 180 deg

So your solution is RIGHT!!
Nothing to add, and sorry to have confused you in my previous solution :)

