Finding Angles A and B for Triangular Fire Spotting

  • Thread starter Thread starter morr485
  • Start date Start date
  • Tags Tags
    Angles Fire
AI Thread Summary
The discussion focuses on calculating angles A and B in a triangle formed by two towers, A and B, which are 18.5 miles apart. The bearings from tower A to B is N65E, while the fire's bearings from A and B are N28E and N16.5W, respectively. A user describes a method to visualize the problem by drawing the relevant angles and lines. They determine that angle A is 37 degrees, derived from the difference between the bearings. For angle B, the calculation involves using the sum of angles in a straight line, leading to the conclusion that angle B can be found by subtracting the other angles from 180 degrees.
morr485
Messages
9
Reaction score
0
1. Two towers A and B are 18.5 miles apart. The bearing from A to B is N65E. A fire is spotted
by ranger at both towers. Its bearing from A and B are N28e and N16.5W. This makes a
triangle with A and B and C the fire.
Can anyone give me a clue and finding angles A and B?2. a/sin A = b/sin B
3. I found angle B was 138.5, except it wasn't right.
 
Last edited:
Physics news on Phys.org
I'm trying to post my question.
 
morr485 said:
1. Two towers A and B are 18.5 miles apart. The bearing from A to B is N65E. A fire is spotted
by ranger at both towers. Its bearing from A and B are N28e and N16.5W.
Can anyone give me a clue and finding angles A and B?


2. a/sin A = b/sin B



3. I found angle B was 138.5, except it wasn't right.

Mark a point "A" on your paper. Draw a vertical line (representing north) and a horizontal line (representing east/west). Draw a line slanting 65 degrees to the right of the upward vertical (N65E)and mark its end "B". Draw a line from A slanting 28 degrees to the right of the upward vertical (N28E) and draw a line from B slanting 16.5 degrees to the left of the upward vertical (N16,5W). The fire is at the point where the two lines meet and it and A and B form the triangle you want.

Now you should be able to see that the angle at A is the difference 65- 28= 37 degrees inside the triangle. At B, the lilne from A makes an angle of 65 degrees with the vertical (opposite interior angles with parallel lines) and that, the 16.5 degree angle, and the angle inside the triangle make make a straight,vertical, line. Their sum must be 180 degrees so the angle inside the triangle, at B, is 180- 65- 16.5 degrees.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top