You are working as a fire ranger and you spot a fire that is 15 km away from your tower. Turning an angle of 21° you can see the next tower, 20 km directly east from you.

Question

Fire Ranger During the summer months, British Columbia experiences numerous forest fires. You are working as a fire ranger in the interior of British Columbia and you spot a fire that is 15 km away from your tower. Turning an angle of 21° you can see the next tower, 20 km directly east from you. Your friend is working at that tower and can also see the fire.

a) Should you or your friend radio in the fire spotted to your respective fire stations? Justify your answer and include a diagram with your solution.

b) Your friend decides to dispatch a water plane to the fire. Your station is not equipped with water 

planes. At what angle should the plane fly to target the fire accurately (how many degrees north or south of a western direction)?

For every solution, you must provide the following:

 A diagram showing how the given information is used

 A full solution, showing all work

 Any justification/explanation that may be required

 A final statement that answers the given question

please explain thoroughly

Thank you

Answer 1

In all word problems, my first step is to draw a diagram of the problem:

                                                                      Fire

            15 km

             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:

                                                                       C

            15 km

             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