Answer:
[tex]\left|\stackrel{\to}R\right| \approx 37.84\text{ mph}[/tex]
[tex]\theta_R \approx 26.56\° \ NW[/tex]
Step-by-step explanation:
First, we can solve for the x-component (east-west component) of the wind vector:
[tex]\cos(20\°) = \dfrac{W_x}{18}[/tex]
↓ multiplying both sides by 18
[tex]W_x = 18\cos(20\°)[/tex]
↓ evaluating using a calculator
[tex]W_x \approx 16.91 \text{ mph}[/tex]
Next, we can solve for the y-component (north-south component) of the wind vector:
[tex]\sin(20\°) = \dfrac{W_y}{18}[/tex]
↓ multiplying both sides by 18
[tex]W_y = 18\sin(20\°)[/tex]
↓ evaluating using a calculator
[tex]W_y \approx 6.16\text{ mph}[/tex]
__________________________
Using these components, we can solve for the x- and y-components of the resultant vector.
Since the wind is the only force in the x direction, its x-component is equal to that of the resultant vector:
[tex]R_x = 16.91 \text{ mph}[/tex]
In the y-direction, the two forces are: the pigeon's flight and the y-component of the wind. Added together, we get the y-component of the resultant vector:
[tex]R_y = 40 - 6.16[/tex]
[tex]R_y = 33.84\text{ mph}[/tex]
(Notice how the wind is subtracted due to being in the opposite direction.)
Finally, we can get the magnitude of the resultant vector using the Pythagorean Theorem:
[tex]c^2 = a^2 + b^2[/tex]
↓ replacing with the relevant variables
[tex]\left|\stackrel{\to}R\right|^2 = (R_x)^2 + (R_y)^2[/tex]
↓ taking the square root of both sides
[tex]\displaystyle \left|\stackrel{\to}R\right| = \sqrt{(R_x)^2 + (R_y)^2}[/tex]
↓ plugging in the known values
[tex]\left|\stackrel{\to}R\right| =\sqrt{33.84^2 + 16.91^2}[/tex]
↓ evaluating using a calculator
[tex]\boxed{\left|\stackrel{\to}R\right| \approx 37.84\text{ mph}}[/tex]
And we can solve for the bearing using inverse trigonometry:
[tex]\theta_R = \tan^{-1}\left(\dfrac{R_x}{R_y}\right)[/tex]
↓ plugging in the known values
[tex]\theta_R = \tan^{-1}\left(\dfrac{16.91}{33.84}\right)[/tex]
↓ evaluating using a calculator
[tex]\boxed{\theta_R \approx 26.56\° \ NW}[/tex]
