two radar stations A and B are 80km apart and B is due east of A.One aircraft is on a bearing 030 from A and 346 fromB. A second aircraft is on a bearing of 325 from A and 293 from B.How far apart are the two aircraft?
how slopes
Bye
How can we do through trigonometry while drawing scales
To find the distance between the two aircraft, we can use the concept of bearing triangulation.
Let's first find the distance of each aircraft from the radar station A using the bearing information.
For the first aircraft:
1. The bearing of 030 from A means it is 30 degrees clockwise from the north direction.
2. Since B is due east of A, the bearing from B to the aircraft will be the same as the bearing from A to the aircraft.
3. The bearing of 346 from B is 346 degrees clockwise from the north direction.
4. To find the bearing of the aircraft from B, we subtract the bearing of B from the bearing of the aircraft (346 - 30 = 316 degrees clockwise from the north direction).
Now we have the bearings of the first aircraft from both A (030) and B (316). We can use trigonometry to find the distances.
Using the Law of Sines, we have:
sin(Angle A) / Distance A = sin(Angle B) / Distance B
For the first aircraft:
sin(30) / Distance A = sin(316) / Distance B
We can rearrange this equation to find Distance A:
Distance A = (sin(30) * Distance B) / sin(316)
Now we can calculate the distance of the first aircraft from A.
For the second aircraft:
Similarly, we can find the distance of the second aircraft from A using its bearing information.
Distance A = (sin(325) * Distance B) / sin(293)
Now that we have the distances of both aircraft from A, we can find the distance between the two aircraft by subtracting the distances.
Distance between the two aircraft = |Distance A (first aircraft) - Distance A (second aircraft)|
Hope this helps!
wow - this is a bit different from the usual ones.
Set A at (0,0). Then B is at (80,0).
Label the two aircraft C and D. We want to find CD.
AC has slope tan60° = 1.732
BC has slope tan104° = -4.011
AD has slope tan125° = -1.428
BD has slope tan157° = 0.424
So, C is at the intersection of
y = 1.732x
y = -4.011(x-80)
D is at the intersection of
y = -1.428x
y = 0.424(x-80)
So, solve for C and D, then use the normal distance formula.