Two planes leave the same airport. The first plane leaves at 4.45 pm and averages 484 mph at a bearing of S61°E. The second plane leaves at 5.00 pm and averages 428 mph at a bearing of N19°W.

Question

a) How far apart are the planes at 6.15 pm?

b) What is the bearing from the first plane to the second plane at that time?

Answer 1

First, you should draw a diagram showing each of the planes leaving the airport. I used the origin of a graph (0,0). The first plane went southeast. According to the bearing given, S61°E, the angle from the y axis should be 61° (to the east or right), while the second should be 19° from the y-axis (to the west or left). This will essentially give us 2 right triangles (triangle 1 for plane 1 and triangle 2 for plane 2), where the hypotenuse of each triangle can be calculated using the information about the times and speeds (plane 1 traveled from 4:45 pm to 6:15 pm at a speed of 484 mph; plane 2 traveled from 5:00 pm to 6:15 pm at a speed of 428 mph).

plane 1: time = 1.5 hours 484 mph • 1.5 hours = 726 miles

plane 1: time = 1.25 hours 428 mph • 1.25 hours = 535 miles

From here, we can determine the other parts of our triangles (angles and sides). The angles are the easiest. Since these are right triangles, and we know a second angle for each, we use the fact that all three angles will sum to 180°.

triangle 1: 180° - 90° - 61° = 29°

triangle 2 1: 180° - 90° - 19° = 71°

And the 2 other sides can be determined using the law of sines, a / (sin (A)) = b / (sin (B)) = c / (sin (C)), and the pythagoreon theorem (a² + b² = c²).

triangle 1: a = (726) • ((sin(61)) / (sin (90)) = 635 635² + b² = 726² b = 352

triangle 2: a = (535) • ((sin(19)) / (sin (90)) = 174 174² + b² = 535² b = 506

Now, considering that I put this on graph paper, I can use the coordinates to determine the distance the planes are from each other.

plane 1 coordinates (635, -352)

plane 2 coordinates (-174, 506)

This means that the difference in the x values is 809, while the difference in the y values is 858. This again gives us a right triangle (triangle 3) where the hypotenuse is equal to the distance (in miles) between the two planes.

triangle 3: a² + b² = c² 858² + 809² = c² c = 1179

From here, we need to determine the angles of triangle 3. Back to the law of sines.

triangle 3: sin(A) = 858 / 1179 (sin(90°)) sin (A) = 0.7277 sin^-1(0.7277) = 46.7°

Now, from the perspective of the first plane, the second plane is northwest of it. To determine the bearing, we draw one more right triangle, and we see since the fourth triangle and the third triangle make a right angle (90°), so the angles are complementary. This allows us to solve for that angle.

triangle 4: 90° - 46.7° = 43.3°. This gives us a bearing of N43.3°W