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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: enter image description here

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  • $\begingroup$ could you please upload the drawing of this situation that you came up with? $\endgroup$
    – chouaib
    Nov 18, 2014 at 2:23

1 Answer 1

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EDIT

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

  • First:

enter image description here

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: enter image description here

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 :)

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  • $\begingroup$ I just uploaded a picture of what the situation looks like. And please be aware that the angles on the interior of the triangle were calculated and added by me. They weren't given. $\endgroup$ Nov 18, 2014 at 2:31
  • $\begingroup$ @Cherry_Developer: yes I see that, and I'm half-convinced that your solution is right...I can say that there are some ambiguities here, did you find this exercise in a book ? $\endgroup$
    – chouaib
    Nov 18, 2014 at 2:34
  • $\begingroup$ yes I did. Would you like to see that too? $\endgroup$ Nov 18, 2014 at 2:35
  • $\begingroup$ If you don't mind :) I was hoping it is a given task by your teacher, so you can turn back and ask for additional clarifications. OK let's try ;) $\endgroup$
    – chouaib
    Nov 18, 2014 at 2:41
  • $\begingroup$ Ok. I took a picture and uploaded it. $\endgroup$ Nov 18, 2014 at 2:44

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