Answer:
Approximately [tex]26.6^{\circ}[/tex] north of west.
Step-by-step explanation:
Let the first component ([tex]x[/tex]-component) denote motions in the east-west direction, with east being positive. Let the second component ([tex]y[/tex]-component) denote motions in the north-south direction with north being positive.
The velocity vector of the pigeon relative to the wind would be:
[tex]\displaystyle \begin{bmatrix}0 \\ 40\end{bmatrix}[/tex].
The velocity of the wind relative to the ground is at [tex]20^{\circ}[/tex] south of west. Refer to the diagram attached. When measured in the counterclockwise direction, the angle between this vector and the positive [tex]x[/tex]-axis would be [tex](20 + 180)^{\circ} = 200^{\circ}[/tex].
The velocity vector of the wind would be:
[tex]\displaystyle 18\, \begin{bmatrix}\cos((180 + 20)^{\circ}) \\ \sin((180 + 20)^{\circ})\end{bmatrix}[/tex].
The velocity of the pigeon relative to the ground is the sum of the following two:
- velocity of pigeon relative to the wind, and
- velocity of wind relative to the ground.
[tex]\begin{aligned} \begin{bmatrix}0 \\ 40\end{bmatrix} + 18\, \begin{bmatrix}\cos((180 + 20)^{\circ}) \\ \sin((180 + 20)^{\circ})\end{bmatrix} \approx \begin{bmatrix} -16.9 \\ 33.8\end{bmatrix}\end{aligned}[/tex],
The magnitude of this vector is:
[tex]\displaystyle \sqrt{(-16.9)^{2} + (33.8)^{2}} \approx 37.8[/tex].
The first component (east-west) of this velocity is negative while the second component (north-south) is positive. Hence, this velocity vector would point between west (negative direction of the first component) and north (positive direction of the second component.)
The direction of the wind velocity is given as an angle between this vector and the east-west direction. To represent the ground velocity of the pigeon in a similar way, it will be necessary to find the angle [tex]\theta[/tex] between this vector and the east-west direction. To do so, take the inverse-tangent of the ratio between the absolute value of velocity in the north-south direction (second component, opposite to the required angle) and the absolute value of velocity in the east-west direction (first component, adjacent to the required angle.)
[tex]\begin{aligned} \tan(\theta) &= \frac{(\text{opposite})}{(\text{adjacent})} \\ &= \frac{(\text{second component})}{(\text{first component})} \\ &\approx \frac{16.9}{33.8} \end{aligned}[/tex].
[tex]\begin{aligned} \theta &= \arctan\left(\frac{\text{opposite}}{\text{adjacent}}\right) \\ &\approx \arctan\left(\frac{16.9}{33.8}\right) \\ &\approx 26.6^{\circ}\end{aligned}[/tex].
In other words, the ground velocity of this pigeon would be at a direction of approximately [tex]26.6^{\circ}[/tex] north of west.
