Villages A,B,C and D are such that B is 4km due east A,C is 3km due south of B and D is 4km s50 degrees W from C calculate the distance and bearing of A from D
To find the distance and bearing of A from D, we can use the law of cosines to find the distance between C and D and the law of sines to find the bearing.
First, let's draw a diagram:
```
D
/
/ 4 km
/
C 3 km
|
|
B - 4 km -> A
```
Using the Pythagorean theorem, we can find the distance between C and D:
CD² = BC² + BD²
CD² = 3² + 4²
CD² = 9 + 16
CD² = 25
CD = √25
CD = 5 km
Now, let's find the bearing of A from D:
sin(θ) = BC / CD
sin(θ) = 3 / 5
θ = sin⁻¹(3 / 5)
θ ≈ 36.87°
Therefore, the distance between A and D is 5 km and the bearing of A from D is approximately 36.87°.
call E at 4 cos 50 south of C
CE = 4 cos 50 = 2.57
so BE = 5.57
call F = west to east line with F 5.57 straight down from A
FD = 4 - 4 sin 50 = 4 - 3.06 = 0.94
tan angle FDA = 5.93
so angle FDA = 80.4
so angle FAD = 9.6 deg `
angle of DA clockwise from north = 360 - 9.6 = 350.4 degrees true
AD = 5.57 / cos 9.6 = 5.65
Correcting for the errors in the previous response:
Given the information:
- CE = 4 cos 50° = 2.57 km
- BE = BC + CE = 3 km + 2.57 km = 5.57 km
- FD = 4 km - 4 sin 50° = 4 km - 3.06 km = 0.94 km
- tan angle FDA = 5.93 (this value is incorrect)
The correct calculations are as follows:
Angle FDA (angle between F, D, and A):
tan(angle FDA) = FD / BE
tan(angle FDA) = 0.94 km / 5.57 km
angle FDA ≈ 9.68°
Angle FAD (angle from F to A through D):
angle FAD = 180° - angle FDA
angle FAD = 180° - 9.68°
angle FAD ≈ 170.32°
Angle of DA (angle clockwise from north):
angle of DA = 360° - angle FAD
angle of DA = 360° - 170.32°
angle of DA ≈ 189.68°
AD (distance between A and D):
AD = BE / cos(angle FAD)
AD = 5.57 km / cos(170.32°)
AD ≈ 5.65 km
Therefore, the corrected values are:
- Angle of DA (clockwise from north): 189.68° (rounded to two decimal places)
- AD (distance between A and D): 5.65 km (rounded to two decimal places)
They asked for the bearing of A from D, not of D from A
A is just a little west of due north of D
almost 360 degrees on my pelorus.
Apologies for the confusion. To calculate the bearing of A from D, we can use the corrected information provided:
The angle of DA (angle clockwise from north) is approximately 189.68°. However, since we are looking for the bearing of A from D, we need to find the bearing relative to D.
To determine the bearing of A from D, subtract the angle of DA from 360°:
Bearing of A from D = 360° - 189.68° ≈ 170.32°
Therefore, the bearing of A from D is approximately 170.32°, indicating that A is just a little west of due north from D.
a little west of north is almost 360 ( navigation teacher here)
170 is almost SOUTH
LOOK, you went east, then south, then southwest
To head back you will have to go a trifle west of due NORTH
That is about 0 or 360, not close to South which is 180
Apologies for the confusion in the previous responses. You are correct.
To clarify, the correct bearing of A from D, considering A being a little west of due north of D, is indeed close to 0 or 360 degrees.
Therefore, the bearing of A from D is approximately 0 or 360 degrees. A is a little west of due north of D.