The bearing of the plane is approximately 178.037°. ![\blacksquare](https://tex.z-dn.net/?f=%5Cblacksquare)
<h3>Procedure - Determination of the bearing of the plane</h3><h3 />
Let suppose that <em>bearing</em> angles are in the following <em>standard</em> position, whose vector formula is:
(1)
Where:
- Magnitude of the vector, in miles per hour.
- Direction of the vector, in degrees.
That is, the line of reference is the
semiaxis.
The <em>resulting</em> vector (
), in miles per hour, is the sum of airspeed of the airplane (
), in miles per hour, and the speed of the wind (
), in miles per hour, that is:
(2)
If we know that
,
,
and
, then the resulting vector is:
![\vec v = 239 \cdot (\sin 180^{\circ}, \cos 180^{\circ}) + 10\cdot (\sin 53^{\circ}, \cos 53^{\circ})](https://tex.z-dn.net/?f=%5Cvec%20v%20%3D%20239%20%5Ccdot%20%28%5Csin%20180%5E%7B%5Ccirc%7D%2C%20%5Ccos%20180%5E%7B%5Ccirc%7D%29%20%2B%2010%5Ccdot%20%28%5Csin%2053%5E%7B%5Ccirc%7D%2C%20%5Ccos%2053%5E%7B%5Ccirc%7D%29)
![\vec v = (7.986, -232.981) \,\left[\frac{mi}{h} \right]](https://tex.z-dn.net/?f=%5Cvec%20v%20%3D%20%287.986%2C%20-232.981%29%20%5C%2C%5Cleft%5B%5Cfrac%7Bmi%7D%7Bh%7D%20%5Cright%5D)
Now we determine the bearing of the plane (
), in degrees, by the following <em>trigonometric</em> expression:
(3)
![\theta = \tan^{-1}\left(-\frac{7.986}{232.981} \right)](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%5Cleft%28-%5Cfrac%7B7.986%7D%7B232.981%7D%20%5Cright%29)
![\theta \approx 178.037^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%5Capprox%20178.037%5E%7B%5Ccirc%7D)
The bearing of the plane is approximately 178.037°. ![\blacksquare](https://tex.z-dn.net/?f=%5Cblacksquare)
To learn more on bearing, we kindly invite to check this verified question: brainly.com/question/10649078
We can start out simple, so here's some:
If you ever drew a sea saw as a child, or seen it, you probably had seen a part that looks like a triangular prism. It's somewhere in the structure.
Many building infrastructures have them. The roofs of many houses are pointed in some what of a triangle like shape so that rain water can roll off of them easier.
Hope these two help or at least spark some ideas!
The answer would be correct. hope that helped
Based on the weight and the model that is given, it should be noted that W(t) in radians will be W(t) = 0.9cos(2πt/366) + 8.2.
<h3>
How to calculate the radian.</h3>
From the information, W(t) = a cos(bt) + d. Firstly, calculate the phase shift, b. At t= 0, the dog is at maximum weight, so the cosine function is also at a maximum. The cosine function is not shifted, so b = 1.
Then calculate d. The dog's average weight is 8.2 kg, so the mid-line d = 8.2. W(t) = a cos t + 8.2. Then calculate a, the dog's maximum weight is 9.1 kg. The deviation from the average is 9.1 kg - 8.2 kg = 0.9 kg. W(t) = 0.9cost + 8.2
Lastly, calculate t. The period p = 2π/b = 2π/1 = 2π. The conversion factor is 1 da =2π/365 rad. Therefore, the function with t in radians is W(t) = 0.9cos(2πt/365) + 8.2.
Learn more about radians on:
brainly.com/question/12939121