Law of Sines

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mathxyz

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An airplane traveling at an altitude of 2050 feet sights the top of a 50-foot tower at an angle of depression of 28 degrees from point A. After the airplane continues in level flight to point B, the angle of depression to the same tower is 34 degrees. Find, to the nearest foot, the distance that the plane has traveled from point A to point B.

I know that X = distance from point A to point B. This is what I need to find.

How can I use the law of sines to find X?
 
Start with a definition. Great!

By drawing the lines of sight and the path of the aircraft, you construct a triangle. If you also construct a vertical line segment from the tower, perpendicular to, and intersecting the level flight path, you will create an altitude of the triangle. The altitude of the triangle is 2050 ft - 50 ft = 2000 ft. The angles of depression MAY BE the internal angles of the triangle. You have to figure out the angles in the triangle. They may be supplements or compliments or the given angle of depression, depending on how you have drawn things.

Note: This is an extremely poorly worded problem statement.

1) Airplanes don't sight anything.
2) It isn't clear where the aircraft is in relation to the tower. If it keeps flying, will it pass over the tower or is it looking out a starboard porthole?
3) It isn't clear if, or that, the two measurements are taken on the same side of the tower. Did it pass over the tower before the second measurement?
4) It doesn't say if the measurement is to the top of the tower or to the bottom. I assumed the top, above.

Normally, the wording would be, "sights a tower directly in front of the aircraft". This establishes that they are in the same plane.

Normally, the wording should be, "before passing over the tower" or, "after passing over the tower". This establishes a unique solution.

For now, if we assume they are in the same plane, you have two solutions. The ones on the same side require a little algebra. The ones on opposite sides require only direct calculation and addition. It looks like tangents do us more good than sines or cosines.
 
ok

Your notes are very helpful and clear. I like comparing tutor notes.
 
First our picture.
Draw a short vertical line representing the 50 foot high tower, labeling the top of the tower T, and another longer vertical line to the right of the tower representing the inititial location of the plane at 2050 feet altitude.
Assuming the pilot's sightings are straight ahead while the airplane is on a level flight path:
The first sighting from point A has an angle of depression of 28º to the top of the 50 foot tower T.
The second sighting from point B, to the left of point A, has an angle of depression of 34º.
We then have triangle TBA.
With the aid of the 34º depression angle, angle TBA is equal to 146º.
Draw a horizontal line from point T to the line representing the plane's initial altitude point, calling it C (below point A).
In order to gain any dimensional information about triangle TBA, we must first find the distance TA.
From the figure, TA = AC/sin28º = 2000/sin28º.
Having TA, angle TBA = 146º and angle BTA = 6º, the application of the Law of Sines will give you the distance AB.
Good luck.
 
Geeezzzz...
If some dingdong teacher is attempting to show a student how to
apply the Law of Sines, why not simply draw a damn triangle, insert
the necessary "givens", and end up with a clear problem...
why word it such that understanding the damn thing is 99% of the work?
 
...and who said there was no passion in mathematics...
 
passion? one letter too many: should be pass on
(my wish for many math teachers)
 
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