# 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:

• could you please upload the drawing of this situation that you came up with? Nov 18, 2014 at 2:23

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