In all word problems, my first step is to draw a diagram of the problem:
21 ° My tower 20 km Friend's tower
Since I have the dimensions of two sides of the triangle and the angle between them, I can use the cosine rule to calculate other features of the triangle. To make it easier, I will relabel the vertices with letters:
21 ° A 20 km B
According to this diagram, side a (opposite vertex A) is 15 km, side c (opposite vertex C) is 20 km, and side b (opposite vertex B) is unknown. I therefore use a version of the cosine rule with b as the unknown:
a = sqrt ( b2 + c2 - 2bc cos(A) )
Substituting the values, I have a = sqrt ( 152 + 202 - 2*15*20 *cos(21))
Simplifying the terms, I have a = sqrt (225 + 400 - 600 * 0.934)
Simplifying further, I have a = sqrt (625 - 560.148) = sqrt (64.852) = 8.053.
Using a reasonable level of accuracy, we have found that the fire is 8 km from my friend's tower.
We can now use the cosine rule to find the missing two angles in the triangle, although we only actually need angle B. b = sqrt ( a2 + c2 - 2ac cos(B) )
Substituting our values, we have 15 = sqrt ( 82 + 202 - 2*8*20*cos(B))
Simplifying the terms, we have 15 = sqrt (64 + 400 - 320*cos(B))
Squaring both sides, we have 225 = 464 - 320*cos(B)
Subtracting 464 from both sides, we have -239 = -320*cos(B)
Dividing both sides by -320, we have cos(B) = -239/-320 = 0.747
We can now find the arccosin which is 41.7 degrees.
Now, back to the word problem:
Since the fire is 15 km from my tower, and only 8 km from my friend's tower, my friend should be the one to report it.
Assuming the plane will fly directly over my friend's tower on its way to the fire, it should fly 42 degrees north of east.
As a check of my calculations, because the last person I want to find any mistakes is the teacher, so I, myself, will try to find any mistakes first, I will find the angle at the fire.
c = sqrt ( a2 + b2 - 2ab cos(C) )
Substituting our values, we have 20 = sqrt ( 82 + 152 - 2*8*15*cos(C))
Simplifying the terms, we have 20 = sqrt (64 + 225 - 240*cos(B))
Squaring both sides, we have 400 = 289 - 240*cos(B)
Subtracting 289 from both sides, we have 111 = -240*cos(B)
Dividing both sides by -240, we have cos(B) = 111/-240 = -0.4625
We can now find the arccosin which is 117.5 degrees.
Adding our three angle measures, we get 21 + 41.7 + 117.5 = 180.2 degrees which is close enough to 180 degrees (being the sum of the angles in a triangle) considering we have rounded our values. So we can be reassured that our solution is correct