Can u help me with this bearing problem?

Question

Two planes take off at the same time from an airport. The first plan is flying at 260 miles per hour on a bearing of S 45dgree E. The second plane is flying in the direction S 45 degree W at 265 miles per hour. If there are no wind currents blowing, how far apart are they after 3 hours? Please round the answer to the nearest whole number.

____ miles

What is the bearing of the second plane from the first after 3 hours?

___ ___ degree ___

Please explain how to do it too!

The bearing answer is S 89 degree W.

Answer 1

Distance = Velocity / time.

Plane 1:

260/3 = 86.666666667mi to the southeast

Plane 2:

265/3 = 88.33333333mi to the southwest.


Now, make a triangle. It should be a right angle and the above lengths match the legs. The hypotenuse should give you the distance between the two planes. Using the pythagorean theorem, a^2+b^2=c^2


86.666667^2+88.3333333^2= (distance between planes)^2

Make sure to take that square root when solving. (I think it should come out to 123.749mi) After 3 hours of travel, I think that the 2nd plane should still be at a 90 degree angle. It must be a decimal that's very close but rounds down to 89. Not sure on that part. Best of luck.

Answer 2

plane A has speed of 260 mi/hr
plane B has speed of 265 mi/hr
After 3 hours,
plane A has traveled the distance 780 mi
plane B has traveled the distance 795 mi

Since
plane A travels S45°E
plane B travels S45°W
then imagine plane A travels from an origin down SE
and imagine plane B travels from the same origin down SW

You will see that the included angle measures 45° + 45° = 90°

Therefore, the triangle they form is a right triangle, where the legs are the distances that the 2 planes have traveled, and the hypotenuse serves as the distance between them after 3 hours. We can solve for the distance(which is the hypotenuse) using the Pythagorean Theorem.(note: for other cases, you will use the cosine rule to determine the distance-- this case is just a special one)

Let d = distance
d² = (780 mi)² + (795 mi)²
d = 1113.74 mi

The bearing of the second plane from the first is.
180° - 45° - arctan (795/780)
= 89.45°

Therefore, the distance is 1113.75 mi and the bearing from 1st to 2nd plane is S89.45° W

^_^

Answer 3

As the first plane, Plane A is travelling SOuth East at 260miles per hour and the second plane, Plane b is travelling at 265miles per hour, the relative velocity of B with respect to A is,
Using pythargoras,
= ((260^2) + (265^2)) ^ .5
= 371.248
Therefore, 3 hours later, the distance between the two planes is = 371.248 * 3
= 1113.744
= 1114 Miles (to the neasrest miles)
Now to understand the bearing of the second plane from the first one after three hours, draw a right-angled triangle labelling the shorter side 260 miles per hour and the longer side 265 miles per hour. Measure the angle between the 260 miles per hour and the hypotenuse - it can be done using trigonometry,
(assuming that the angle is x),
tan x = 265 / 260
or, x = 45.5 degrees
Now draw the cross pointing at each of the four direction on each of the four point of your triangle, labelling the corner between hypotenuse and 260mph as A, the one between 265mph and hypotenuse as B and the last point as C.
label x = 45.5 degrees.
Extend the East from Point C and the NOrth from Point A so that they meet and note that they form a right-angle isosceles triangle with angles 90, 45 and 45 degrees. Now, measure the bearing of point B from A, which is = 360 - 45 - 45.5 = 269.5 from North - which means the bearing is 89.5 W from S (or South).

Answer 4

Distance = Velocity * time

So after 3 hours : plane1 is 260*3 = 780 miles
plane2 is 265*3= 795 miles
As ballerina said they form a right triangle and we must find the hypothenuse.
x=sqrt(780^2 + 795^2)
x=1113.7 miles or 1114 miles rounded.

The planes are 90 degrees apart.

Answer 5

here is a hint, the flightpaths form two sides of a triangle. Draw a picture it will help you understand.