Third leg.
The crew flies at a speed of 560 mi/h in direction N-20°-E.
The wind has a speed of 35 mi/h and a direction S-10°-E.
We then can draw this as:
We have to add the two vectors to find the actual speed and direction.
We will start by adding the x-coordinate (W-E axis):
![\begin{gathered} x=560\cdot\sin (20\degree)+35\cdot\sin (10\degree) \\ x\approx560\cdot0.342+35\cdot0.174 \\ x\approx191.53+6.08 \\ x\approx197.61 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%3D560%5Ccdot%5Csin%20%2820%5Cdegree%29%2B35%5Ccdot%5Csin%20%2810%5Cdegree%29%20%5C%5C%20x%5Capprox560%5Ccdot0.342%2B35%5Ccdot0.174%20%5C%5C%20x%5Capprox191.53%2B6.08%20%5C%5C%20x%5Capprox197.61%20%5Cend%7Bgathered%7D)
and the y-coordinate (S-N axis) is:
![\begin{gathered} y=560\cdot\cos (20\degree)-35\cdot\cos (10\degree) \\ y\approx560\cdot0.940-35\cdot0.985 \\ y\approx526.23-34.47 \\ y\approx491.76 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y%3D560%5Ccdot%5Ccos%20%2820%5Cdegree%29-35%5Ccdot%5Ccos%20%2810%5Cdegree%29%20%5C%5C%20y%5Capprox560%5Ccdot0.940-35%5Ccdot0.985%20%5C%5C%20y%5Capprox526.23-34.47%20%5C%5C%20y%5Capprox491.76%20%5Cend%7Bgathered%7D)
Then, the actual speed vector is v3=(197.61, 491.76).
The starting location for the third leg is R2=(216.66, 167.67) [taken from the previous answer].
Then, we have to calculate the displacement in 20 minutes using the actual speed vector.
We can calculate the movement in each of the axis. For the x-axis:
![\begin{gathered} R_{3x}=R_{2x}+v_{3x}\cdot t \\ R_{3x}=216.66+197.61\cdot\frac{1}{3} \\ R_{3x}=216.66+65.87 \\ R_{3x}=282.53 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20R_%7B3x%7D%3DR_%7B2x%7D%2Bv_%7B3x%7D%5Ccdot%20t%20%5C%5C%20R_%7B3x%7D%3D216.66%2B197.61%5Ccdot%5Cfrac%7B1%7D%7B3%7D%20%5C%5C%20R_%7B3x%7D%3D216.66%2B65.87%20%5C%5C%20R_%7B3x%7D%3D282.53%20%5Cend%7Bgathered%7D)
NOTE: 20 minutes represents 1/3 of an hour.
We can do the same with the y-coordinate:
![\begin{gathered} R_{3y}=R_{2y}+v_{3y}\cdot t \\ R_{3y}=167.67+491.76\cdot\frac{1}{3} \\ R_{3y}=167.67+163.92 \\ R_{3y}=331.59 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20R_%7B3y%7D%3DR_%7B2y%7D%2Bv_%7B3y%7D%5Ccdot%20t%20%5C%5C%20R_%7B3y%7D%3D167.67%2B491.76%5Ccdot%5Cfrac%7B1%7D%7B3%7D%20%5C%5C%20R_%7B3y%7D%3D167.67%2B163.92%20%5C%5C%20R_%7B3y%7D%3D331.59%20%5Cend%7Bgathered%7D)
The final position is R3 = (282.53, 331.59).
To find the distance from the origin and direction, we transform the cartesian coordinates of R3 into polar coordinates:
The distance can be calculated as if it was a right triangle:
![\begin{gathered} d^2=x^2+y^2_{} \\ d^2=282.53^2+331.59^2 \\ d^2=79823.20+109951.93 \\ d^2=189775.13 \\ d=\sqrt[]{189775.13} \\ d\approx435.63 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20d%5E2%3Dx%5E2%2By%5E2_%7B%7D%20%5C%5C%20d%5E2%3D282.53%5E2%2B331.59%5E2%20%5C%5C%20d%5E2%3D79823.20%2B109951.93%20%5C%5C%20d%5E2%3D189775.13%20%5C%5C%20d%3D%5Csqrt%5B%5D%7B189775.13%7D%20%5C%5C%20d%5Capprox435.63%20%5Cend%7Bgathered%7D)
The angle, from E to N, can be calculated as:
![\begin{gathered} \tan (\alpha)=\frac{y}{x} \\ \tan (\alpha)=\frac{331.59}{282.53} \\ \tan (\alpha)\approx1.1736 \\ \alpha=\arctan (1.1736) \\ \alpha=49.56\degree \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctan%20%28%5Calpha%29%3D%5Cfrac%7By%7D%7Bx%7D%20%5C%5C%20%5Ctan%20%28%5Calpha%29%3D%5Cfrac%7B331.59%7D%7B282.53%7D%20%5C%5C%20%5Ctan%20%28%5Calpha%29%5Capprox1.1736%20%5C%5C%20%5Calpha%3D%5Carctan%20%281.1736%29%20%5C%5C%20%5Calpha%3D49.56%5Cdegree%20%5Cend%7Bgathered%7D)
If we want to express it from N to E, we substract the angle from 90°:
![\beta=90\degree-\alpha=90-49.56=40.44\degree](https://tex.z-dn.net/?f=%5Cbeta%3D90%5Cdegree-%5Calpha%3D90-49.56%3D40.44%5Cdegree)
Answer: the final location can be represented with the vector (282.53, 331.59).
1) The distance from the origin is 435.63 miles and
2) the direction is N-40°-E.