Question 1112902
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*[illustration FireObservers]

Direction angles (typically) are measured clockwise from North, which is typically in the direction of the positive *[tex \Large y]-axis on a coordinate plane.


Thus, if the direction of the fire from A is *[tex \Large 352^{\circ}], then the angle measured on the coordinate plane must be 360 - 352 + 90, that is **[tex \Large 98^{\circ}].


The direction of B from A is *[tex \Large 41^{\circ}], so measured on the coordinate plane the angle is 90 - 41 = 49.


That means that the angle from the line between A and B to the line between A and the Fire must be 98 - 49 = 49 degrees.  Now we have angle A.


Considering a line parallel to the x-axis through point B, measured in the coordinate plane would be 360 - 294 + 90 = 156.  Continuing clockwise, the angle from the line between B and the Fire and the horizonal line through B is 180 - 156 = 24.  By opposite interior angles of a transversal of two parallel lines, the angle from the line between A and B and the horizontal line through B is 49 degrees.  Hence, the angle from the line between A and B and the line between B and the Fire is 24 + 49 = 73 degrees.


Then the final angle of the triangle, at the point of the fire is 180 - (73 + 49) = 66 degrees.


Use the Law of Sines:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{45}{\sin(66)}\ =\ \frac{\overline{AF}}{\sin(73)}\ =\ \frac{\overline{BF}}{\sin(49)}]


The rest is just calculator work.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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